### LEARNING OBJECTIVES

By the end of this section, you will be able to:

- Calculate formula masses for covalent and ionic compounds
- Define the amount unit mole and the related quantity Avogadro’s number
- Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another

We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.

## Formula Mass

In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the **formula mass **of a substance by summing the average atomic masses of all the atoms represented in the substance’s formula.

### Formula Mass for Covalent Substances

For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl_{3}), a covalent compound once used as a surgical anesthetic and now primarily used in the production of the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure 1 outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu.

Likewise, the molecular mass of an aspirin molecule, C_{9}H_{8}O_{4}, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure 2).

#### Example 1

**Computing Molecular Mass for a Covalent Compound**

Ibuprofen, C_{13}H_{18}O_{2}, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?

#### Solution

Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:

**Check Your Learning**

Acetaminophen, C_{8}H_{9}NO_{2}, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?

**Answer**: 151.16 amu

### Formula Mass for Ionic Compounds

Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.”

As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na^{+}, and chloride anions, Cl^{–}, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (see Figure 3).

Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.

### Example 2

**Computing Formula Mass for an Ionic Compound**

Aluminum sulfate, Al_{2}(SO_{4})_{3}, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound?

#### Solution

The formula for this compound indicates it contains Al^{3+} and SO_{4}^{2-} ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al_{2}S_{3}O_{12}. Following the approach outlined above, the formula mass for this compound is calculated as follows:

**Check Your Learning**

Calcium phosphate, Ca_{3}(PO_{4})_{2}, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?

**Answer:**310.18 amu

## The Mole

The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H_{2}O, and hydrogen peroxide, H_{2}O_{2}, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the *mole*, which remains indispensable in modern chemical science.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of *the number* of atoms or molecules in a bulk sample of matter. A **mole** is defined as *the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) as the number of atoms in a sample of pure ^{12}C weighing exactly 12 g.* One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.

The number of entities composing a mole has been experimentally determined to be [latex]6.02214179\times {10}^{23}[/latex], a fundamental constant named **Avogadro’s number** **( N_{A})** or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being [latex]6.022\times {10}^{23}\text{/mol.}[/latex]

Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The** molar mass** of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 4).

Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, ^{12}C, *the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu*. Per the amu definition, a single ^{12}C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of ^{12}C contains 1 mole of ^{12}C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, ^{12}C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 5).

Element | Average Atomic Mass (amu) | Molar Mass (g/mol) | Atoms/Mole |
---|---|---|---|

C | 12.01 | 12.01 | [latex]6.022\times {10}^{23}[/latex] |

H | 1.008 | 1.008 | [latex]6.022\times {10}^{23}[/latex] |

O | 16.00 | 16.00 | [latex]6.022\times {10}^{23}[/latex] |

Na | 22.99 | 22.99 | [latex]6.022\times {10}^{23}[/latex] |

Cl | 33.45 | 33.45 | [latex]6.022\times {10}^{23}[/latex] |

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 6). Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.

The mole is used in chemistry to represent [latex]6.022\times {10}^{23}[/latex] of something, but it can be difficult to conceptualize such a large number. Watch this video to learn more.

The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.

### Example 3

**Deriving Moles from Grams for an Element**

According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?

#### Solution

The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.

The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

**Check Your Learning**

Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?

**Answer**: 0.360 mol

### Example 4

**Deriving Grams from Moles for an Element**

A liter of air contains [latex]9.2\times {10}^{-4}[/latex] mol argon. What is the mass of Ar in a liter of air?

#### Solution

The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10^{–3}) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

The result is in agreement with our expectations as noted above, around 0.04 g Ar.

**Check Your Learning**

What is the mass of 2.561 mol of gold?

**Answer**: 504.4 g

### Example 5

**Deriving Number of Atoms from Mass for an Element**

Copper is commonly used to fabricate electrical wire (Figure 7). How many copper atoms are in 5.00 g of copper wire?

#### Solution

The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (*N _{A}*) to convert this molar amount to number of Cu atoms:

Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth *N _{A}*, or approximately 10

^{22}Cu atoms. Carrying out the two-step computation yields:

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10^{22} as expected.

**Check Your Learning**

A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?

**Answer**: [latex]4.586\times {10}^{22}\text{Au atoms}[/latex]

### Example 6

**Deriving Moles from Grams for a Compound**

Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C_{2}H_{5}O_{2}N. How many moles of glycine molecules are contained in 28.35 g of glycine?

#### Solution

We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 3:

The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C_{2}H_{5}O_{2}N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:

This result is consistent with our rough estimate.

**Check Your Learning**

How many moles of sucrose, C_{12}H_{22}O_{11}, are in a 25-g sample of sucrose?

**Answer**: 0.073 mol

### Example 7

**Deriving Grams from Moles for a Compound**

Vitamin C is a covalent compound with the molecular formula C_{6}H_{8}O_{6}. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is [latex]1.42\times {10}^{-4}\text{mol.}[/latex] What is the mass of this allowance in grams?

#### Solution

As for elements, the mass of a compound can be derived from its molar amount as shown:

The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10^{-4} or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:

This is consistent with the anticipated result.

**Check Your Learning**

What is the mass of 0.443 mol of hydrazine, N_{2}H_{4}?

**Answer**: 14.2 g

### Example 8

**Deriving the Number of Atoms and Molecules from the Mass of a Compound**

A packet of an artificial sweetener contains 40.0 mg of saccharin (C_{7}H_{5}NO_{3}S), which has the structural formula:

Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?

#### Solution

The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Figure 5, and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:

The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:

**Check Your Learning**

How many C_{4}H_{10} molecules are contained in 9.213 g of this compound? How many hydrogen atoms?

**Answer**: [latex]9.545\times {10}^{22}\text{molecules}{\text{C}}_{4}{\text{H}}_{10}\text{;}9.545\times {10}^{23}\text{atoms H}[/latex]

### Counting Neurotransmitter Molecules in the Brain

The brain is the control center of the central nervous system (Figure 8). It sends and receives signals to and from muscles and other internal organs to monitor and control their functions; it processes stimuli detected by sensory organs to guide interactions with the external world; and it houses the complex physiological processes that give rise to our intellect and emotions. The broad field of neuroscience spans all aspects of the structure and function of the central nervous system, including research on the anatomy and physiology of the brain. Great progress has been made in brain research over the past few decades, and the BRAIN Initiative, a federal initiative announced in 2013, aims to accelerate and capitalize on these advances through the concerted efforts of various industrial, academic, and government agencies (more details available at the White House’s website).

Specialized cells called neurons transmit information between different parts of the central nervous system by way of electrical and chemical signals. Chemical signaling occurs at the interface between different neurons when one of the cells releases molecules (called neurotransmitters) that diffuse across the small gap between the cells (called the synapse) and bind to the surface of the other cell. These neurotransmitter molecules are stored in small intracellular structures called vesicles that fuse to the cell wall and then break open to release their contents when the neuron is appropriately stimulated. This process is called exocytosis (see Figure 9). One neurotransmitter that has been very extensively studied is dopamine, C_{8}H_{11}NO_{2}. Dopamine is involved in various neurological processes that impact a wide variety of human behaviors. Dysfunctions in the dopamine systems of the brain underlie serious neurological diseases such as Parkinson’s and schizophrenia.

One important aspect of the complex processes related to dopamine signaling is the number of neurotransmitter molecules released during exocytosis. Since this number is a central factor in determining neurological response (and subsequent human thought and action), it is important to know how this number changes with certain controlled stimulations, such as the administration of drugs. It is also important to understand the mechanism responsible for any changes in the number of neurotransmitter molecules released—for example, some dysfunction in exocytosis, a change in the number of vesicles in the neuron, or a change in the number of neurotransmitter molecules in each vesicle.

Significant progress has been made recently in directly measuring the number of dopamine molecules stored in individual vesicles and the amount actually released when the vesicle undergoes exocytosis. Using miniaturized probes that can selectively detect dopamine molecules in very small amounts, scientists have determined that the vesicles of a certain type of mouse brain neuron contain an average of 30,000 dopamine molecules per vesicle (about [latex]5\times {10}^{-20}[/latex] mol or 50 zmol). Analysis of these neurons from mice subjected to various drug therapies shows significant changes in the average number of dopamine molecules contained in individual vesicles, increasing or decreasing by up to three-fold, depending on the specific drug used. These studies also indicate that not all of the dopamine in a given vesicle is released during exocytosis, suggesting that it may be possible to regulate the fraction released using pharmaceutical therapies.^{[1]}

## Key Concepts and Summary

The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be [latex]6.022\times {10}^{23}[/latex], a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g/mol) for any substance are numerically equivalent (for example, one H_{2}O molecule weighs approximately18 amu and 1 mole of H_{2}O molecules weighs approximately 18 g).

### Chemistry End of Chapter Exercises

- What is the total mass (amu) of carbon in each of the following molecules?
- CH
_{4 } - CHCl
_{3 } - C
_{12}H_{10}O_{6 } - CH
_{3}CH_{2}CH_{2}CH_{2}CH_{3}

- CH
- What is the total mass of hydrogen in each of the molecules?
- CH
_{4 } - CHCl
_{3 } - C
_{12}H_{10}O_{6 } - CH
_{3}CH_{2}CH_{2}CH_{2}CH_{3}

- CH
- Calculate the molecular or formula mass of each of the following:
- P
_{4 } - H
_{2}O - Ca(NO
_{3})_{2 } - CH
_{3}CO_{2}H (acetic acid) - C
_{12}H_{22}O_{11}(sucrose, cane sugar).

- P
- Determine the molecular mass of the following compounds:
- Determine the molecular mass of the following compounds:
- Which molecule has a molecular mass of 28.05 amu?
- Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.
- Compare 1 mole of H
_{2}, 1 mole of O_{2}, and 1 mole of F_{2}.- Which has the largest number of molecules? Explain why.
- Which has the greatest mass? Explain why.

- Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C
_{2}H_{5}OH), 0.60 mol of formic acid (HCO_{2}H), or 1.0 mol of water (H_{2}O)? Explain why. - Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C
_{2}H_{5}OH), 1 mol of formic acid (HCO_{2}H), or 1 mol of water (H_{2}O)? Explain why. - How are the molecular mass and the molar mass of a compound similar and how are they different?
- Calculate the molar mass of each of the following compounds:
- hydrogen fluoride, HF
- ammonia, NH
_{3 } - nitric acid, HNO
_{3 } - silver sulfate, Ag
_{2}SO_{4 } - boric acid, B(OH)
_{3}

- Calculate the molar mass of each of the following:
- S
_{8 } - C
_{5}H_{12 } - Sc
_{2}(SO_{4})_{3 } - CH
_{3}COCH_{3}(acetone) - C
_{6}H_{12}O_{6}(glucose)

- S
- Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals:
- limestone, CaCO
_{3 } - halite, NaCl
- beryl, Be
_{3}Al_{2}Si_{6}O_{18 } - malachite, Cu
_{2}(OH)_{2}CO_{3 } - turquoise, CuAl
_{6}(PO_{4})_{4}(OH)_{8}(H_{2}O)_{4}

- limestone, CaCO
- Calculate the molar mass of each of the following:
- the anesthetic halothane, C
_{2}HBrClF_{3 } - the herbicide paraquat, C
_{12}H_{14}N_{2}Cl_{2 } - caffeine, C
_{8}H_{10}N_{4}O_{2 } - urea, CO(NH
_{2})_{2 } - a typical soap, C
_{17}H_{35}CO_{2}Na

- the anesthetic halothane, C
- Determine the number of moles of compound and the number of moles of each type of atom in each of the following:
- 25.0 g of propylene, C
_{3}H_{6 } - [latex]3.06\times {10}^{-3}\text{g}[/latex] of the amino acid glycine, C
_{2}H_{5}NO_{2 } - 25 lb of the herbicide Treflan, C
_{13}H_{16}N_{2}O_{4}F (1 lb = 454 g) - 0.125 kg of the insecticide Paris Green, Cu
_{4}(AsO_{3})_{2}(CH_{3}CO_{2})_{2 } - 325 mg of aspirin, C
_{6}H_{4}(CO_{2}H)(CO_{2}CH_{3})

- 25.0 g of propylene, C
- Determine the mass of each of the following:
- 0.0146 mol KOH
- 10.2 mol ethane, C
_{2}H_{6 } - [latex]1.6\times {10}^{-3}\text{mol}{\text{Na}}_{2}{\text{SO}}_{4}[/latex]
- [latex]6.854\times {10}^{3}\text{mol glucose},{\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}[/latex]
- 2.86 mol Co(NH
_{3})_{6}Cl_{3}

- Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:
- 2.12 g of potassium bromide, KBr
- 0.1488 g of phosphoric acid, H
_{3}PO_{4 } - 23 kg of calcium carbonate, CaCO
_{3 } - 78.452 g of aluminum sulfate, Al
_{2}(SO_{4})_{3 } - 0.1250 mg of caffeine, C
_{8}H_{10}N_{4}O_{2}

- Determine the mass of each of the following:
- 2.345 mol LiCl
- 0.0872 mol acetylene, C
_{2}H_{2 } - [latex]3.3\times {10}^{-2}\text{mol}{\text{Na}}_{2}{\text{CO}}_{3}[/latex]
- [latex]1.23\times {10}^{3}\text{mol fructose,}{\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}[/latex]
- 0.5758 mol FeSO
_{4}(H_{2}O)_{7}

- The approximate minimum daily dietary requirement of the amino acid leucine, C
_{6}H_{13}NO_{2}, is 1.1 g. What is this requirement in moles? - Determine the mass in grams of each of the following:
- 0.600 mol of oxygen atoms
- 0.600 mol of oxygen molecules, O
_{2 } - 0.600 mol of ozone molecules, O
_{3}

- A 55-kg woman has [latex]7.5\times {10}^{-3}\text{mol}[/latex] of hemoglobin (molar mass = 64,456 g/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?
- Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO
_{4}, a semiprecious stone. - Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH
_{4}, 0.6 mol of C_{6}H_{6}, or 0.4 mol of C_{3}H_{8}. - Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO
_{4}, 266 g of A1_{2}C1_{6}, or 225 g of A1_{2}S_{3}. - Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?
- The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone
- One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?
- A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C
_{12}H_{22}O_{11}) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar? - A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na
_{2}PO_{3}F) in 100 mL- What mass of fluorine atoms in mg was present?
- How many fluorine atoms were present?

- Which of the following represents the least number of molecules?
- 20.0 g of H
_{2}O (18.02 g/mol) - 77.0 g of CH
_{4}(16.06 g/mol) - 68.0 g of CaH
_{2}(42.09 g/mol) - 100.0 g of N
_{2}O (44.02 g/mol) - 84.0 g of HF (20.01 g/mol)

- 20.0 g of H

#### Selected Answers

1. (a) [latex]1\times 12.01\text{amu}=12.01\text{amu;}[/latex] (b) [latex]1\times 12.01\text{amu}=12.01\text{amu;}[/latex] (c) [latex]12\times 12.01\text{amu}=144.12\text{amu;}[/latex] (d) [latex]5\times 12.01\text{amu}=60.05\text{amu}[/latex]

3. (a) [latex]4\times 30.974\text{amu}=123.896\text{amu}[/latex]

(b) [latex]2\times 1.008\text{amu}+15.999\text{amu}=18.015\text{amu}[/latex]

(c) [latex]40\times 0.078\text{amu}+2\times 14.007\text{amu}+6\times 15.999\text{amu}=164.086\text{amu}[/latex]

(d) [latex]2\times 12.011\text{amu}+4\times 1.008\text{amu}+2\times 15.999\text{amu}=60.052\text{amu}[/latex]

(e) [latex]12\times 12.011\text{amu}+22\times 1.008\text{amu}\times 11\times 15.999\text{amu}=342.297\text{amu}[/latex]

5. (a) C_{4}H_{8}

[latex]\begin{array}{ll}4\text{C}\times 12.011\hfill & =48.044\text{amu}\hfill \\ 8\text{H}\times 1.0079\hfill & =\underline{8.06352\text{amu}}\hfill \\ \hfill & =56.107\text{amu}\hfill \end{array}[/latex]

(b) C_{4}H_{6}

(c) H_{2}Si_{2}Cl_{4}

(d) H_{3}PO_{4}

13. (a) S_{8}

[latex]8\text{S}=8\times 32.066=256.528\text{g/mol;}[/latex]

(b) C_{5}H_{12}

[latex]\begin{array}{lll}\hfill 5\text{C}=5\times 12.011& =& 60.055\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{12H}=12\times 1.00794& =& \underline{12.09528\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 72.150\text{g}{\text{mol}}^{-1}\hfill \end{array}\text{;}[/latex]

(c) Sc_{2}(SO_{4})_{3}

[latex]\begin{array}{lll}\hfill 2\text{Sc}=2\times 44.9559109& =& 89.9118218\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{3S}=3\times 32.066& =& 96.198\text{g}{\text{mol}}^{-1}\hfill \\ \hfill 12\text{O}=12\times 15.99943& =& \underline{191.99316\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 378.103\text{g}{\text{mol}}^{-1}\hfill \end{array}\text{;}[/latex]

(d) CH_{3}COCH_{3}

[latex]\begin{array}{lll}\hfill 3\text{C}=3\times 12.011& =& 36.033\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{1O}=1\times 15.9994& =& 15.9994\text{g}{\text{mol}}^{-1}\hfill \\ \hfill 6\text{H}=6\times 1.00794& =& \underline{6.04764\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 58.080\text{g}{\text{mol}}^{-1}\hfill \end{array}\text{;}[/latex]

(e) C_{6}H_{12}O_{6}

[latex]\begin{array}{lll}\hfill \text{6C}=6\times 12.011& =& 72.066\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{12H}=12\times 1.00794& =& 12.09528\text{g}{\text{mol}}^{-1}\hfill \\ \hfill 6\text{O}=6\times 15.9994& =& \underline{95.9964\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 180.158\text{g}{\text{mol}}^{-1}\hfill \end{array}[/latex]

15. (a) C_{2}HBrClF_{3}

[latex]\begin{array}{lll}\hfill \text{2C}=2\times 12.011& =& 24.022\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{1H}=1\times 1.00794& =& 1.00794\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{1Br}=1\times 79.904& =& 79.904\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{1Cl}=1\times 35.453& =& 35.453\text{g}{\text{mol}}^{-1}\hfill \\ \hfill 3\text{F}=3\times 18.998403& =& \underline{56.995209\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 197.682\text{g}{\text{mol}}^{-1}\hfill \end{array}\text{;}[/latex]

(b) C_{12}H_{14}N_{2}Cl_{2}

[latex]\begin{array}{lll}\hfill \text{12C}=12\times 12.011& =& 144.132\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{14H}=14\times 1.00794& =& 14.111\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{2N}=2\times 14.0067& =& 28.0134\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{2Cl}=2\times 35.453& =& \underline{70.906\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 257.163\text{g}{\text{mol}}^{-1}\hfill \end{array}\text{;}[/latex]

(c) C_{8}H_{10}N_{4}O_{2}

[latex]\begin{array}{lll}\hfill \text{8C}=8\times 12.011& =& 96.088\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{10H}=10\times 1.007& =& 10.079\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{4N}=4\times 14.0067& =& 56.027\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{2O}=2\times 15.9994& =& \underline{31.999\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 194.193\text{g}{\text{mol}}^{-1}\hfill \end{array}\text{;}[/latex]

(d) CO(NH_{2})_{2}

[latex]\begin{array}{lll}\hfill \text{1C}=1\times 12.011& =& 12.011\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{1O}=1\times 15.9994& =& 15.9994\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{2N}=2\times 14.0067& =& 28.0134\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{4H}=4\times 1.00794& =& \underline{4.03176\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 60.056\text{g}{\text{mol}}^{-1}\hfill \end{array}\text{;}[/latex]

(e) C_{17}H_{35}CO_{2}Na

[latex]\begin{array}{lll}\hfill \text{18C}=18\times 12.011& =& 216.198\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{35H}=35\times 1.00794& =& 35.2779\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{2O}=2\times 15.9994& =& 31.9988\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{1Na}=1\times 22.98977& =& \underline{22.98977\text{g}{\text{mol}}^{-1}}\hfill \\ \hfill & =& 306.464\text{g}{\text{mol}}^{-1}\hfill \end{array}[/latex]

17. (a) KOH:

[latex]\begin{array}{lll}\hfill 1\text{K}=1\times 39.0983& =& 39.0983\hfill \\ \hfill 1\text{O}=1\times 15.9994& =& 15.9994\hfill \\ \hfill \text{1H}=1\times 1.00794& =& \underline{1.00794}\hfill \\ \hfill \text{molar mass}& =& 56.1056\text{g}{\text{mol}}^{-1}\hfill \end{array}[/latex]

[latex]\text{Mass}=0.0146\text{mol}\times 56.1056\text{g/mol}=0.819\text{g;}[/latex]

(b) C_{2}H_{6}

[latex]\begin{array}{lll}\hfill 2\text{C}=2\times 12.011& =& 24.022\hfill \\ \hfill 6\text{H}=6\times 1.00794& =& \underline{6.04764}\hfill \\ \hfill \text{molar mass}& =& 30.070\text{g}{\text{mol}}^{-1}\hfill \end{array}[/latex]

[latex]\text{Mass}=10.2\text{mol}\times 30.070\text{g/mol}=307\text{g;}[/latex]

(c) Na_{2}SO_{4}:

[latex]\begin{array}{lll}\hfill 2\text{Na}=2\times 22.990& =& 45.98\hfill \\ \hfill 1\text{S}=1\times 32.066& =& 32.066\hfill \\ \hfill 4\text{O}=4\times 15.9994& =& \underline{63.9976}\hfill \\ \hfill \text{molar mass}& =& 142.044\text{g}{\text{mol}}^{-1}\hfill \end{array}[/latex]

[latex]\text{Mass}=1.6\times {10}^{-3}\text{mol}\times 142.044\text{g/mol}=\text{0.23 g;}[/latex]

(d) C_{6}H_{12}O_{6}

[latex]\begin{array}{lll}\hfill 6\text{C}=6\times 12.011& =& 72.066\hfill \\ \hfill 12\text{H}=12\times 1.00794& =& 12.0953\hfill \\ \hfill 6\text{O}=6\times 15.9994& =& \underline{95.9964}\hfill \\ \hfill \text{molar mass}& =& 180.158\text{g}{\text{mol}}^{-1}\hfill \end{array}[/latex]

[latex]\text{Mass}=6.854\times {10}^{3}\text{mol}\times 180.158\text{g/mol}=1.235\times {10}^{6}\text{g (}1235\text{kg);}[/latex]

(e) Co(NH_{3})_{6}Cl_{3}

19. (a) [latex]\begin{array}{lll}\hfill \text{molar mass}\left(\text{LiCl}\right)=1\times 6.941+1\times 35.4527& =& 42.394\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{mass}=2.345\cancel{\text{mol}}\times 42.394\text{g}\cancel{{\text{mol}}^{-1}}& =& 99.41\text{g}\hfill \end{array}[/latex]

(b) [latex]\begin{array}{lll}\hfill \text{molar mass}\left({\text{C}}_{2}{\text{H}}_{2}\right)=2\times 12.011+2\times 1.00794& =& 26.038\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{mass}=0.0872\cancel{\text{mol}}\times 26.038\text{g}\cancel{{\text{mol}}^{-1}}& =& 2.27\text{g}\hfill \end{array}[/latex]

(c) [latex]\begin{array}{lll}\hfill \text{molar mass}\left({\text{Na}}_{2}{\text{CO}}_{3}\right)=2\times 22.989768+1\times 12.011+3\times 15.9994& =& 105.989\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{mass}=3.3\times {10}^{-2}\cancel{\text{mol}}\times 105.989\text{g}{\text{mol}}^{-1}& =& 3.5\text{g}\hfill \end{array}[/latex]

(d) [latex]\begin{array}{lll}\hfill \text{molar mass}\left({\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}\right)=6\times 12.011+12\times 1.00794+6\times 15.9994& =& 180.158\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{mass}=1.23\times {10}^{3}\cancel{\text{mol}}\times 180.158\text{g}{\text{mol}}^{-1}=2.22\times {10}^{5}\text{g}& =& 222\text{kg}\hfill \end{array}[/latex]

(e) [latex]\begin{array}{lll}\hfill \text{molar mass}\left[{\text{FeSO}}_{4}{\left({\text{H}}_{2}\text{O}\right)}_{7}\right]& =& 1\times 55.847+1\times 32.066+4\times 15.999\hfill \\ \hfill +7\left(2\times 1.00794+15.9994\right)& =& 278.018\text{g}{\text{mol}}^{-1}\hfill \\ \hfill \text{mass}=0.5758\cancel{\text{mol}}\times 278.018\text{g}\cancel{{\text{mol}}^{-1}}& =\hfill & 160.1\text{g}\hfill \end{array}[/latex]

21. (a) [latex]0.600\cancel{\text{mol}}\times 15.9994\text{g/}\cancel{\text{mol}}=9.60\text{g}[/latex]

(b) [latex]0.600\cancel{\text{mol}}\times 2\times 15.994\text{g/}\cancel{\text{mol}}=19.2\text{g}[/latex]

(c) [latex]0.600\cancel{\text{mol}}\times 3\times 15.994\text{g/}\cancel{\text{mol}}=28.8\text{g}[/latex]

23. Determine the number of moles of each component. From the moles, calculate the number of atoms and the mass of the elements involved. Zirconium: [latex]0.3384\cancel{\text{mol}}\times 6.022\times {10}^{23}{\cancel{\text{mol}}}^{\cancel{-1}}=2.038\times 1023\text{atoms;}0.3384\cancel{\text{mol}}\times 91.224\text{g/}\cancel{\text{mol}}=30.87\text{g;}[/latex] Silicon: [latex]0.3384\cancel{\text{mol}}\times 6.022\times {10}^{23}{\cancel{\text{mol}}}^{\cancel{-1}}=2.038\times {10}^{23}\text{atoms;}0.3384\cancel{\text{mol}}\times 28.0855\text{g/}\cancel{\text{mol}}=9.504\text{g;}[/latex] Oxygen: [latex]4\times 0.3384\cancel{\text{mol}}\times 6.022\times {10}^{23}{\cancel{\text{mol}}}^{\cancel{-1}}=8.151\times {10}^{23}\text{atoms;}4\times 0.3384\cancel{\text{mol}}\times 15.9994\text{g/}\cancel{\text{mol}}=21.66\text{g}[/latex]

25. Determine the molar mass and, from the grams present, the moles of each substance. The compound with the greatest number of moles of Al has the greatest mass of Al.

Molar mass AlPO_{4}: 26.981539 + 30.973762 + 4(15.9994) = 121.9529 g/mol

Molar mass Al_{2}Cl_{6}: 2(26.981539) + 6(35.4527) = 266.6793 g/mol

Molar mass Al_{2}S_{3}: 2(26.981539) + 3(32.066) = 150.161 g/mol

AlPO_{4}: [latex]\frac{122\cancel{\text{g}}}{121.9529\cancel{\text{g}}{\text{mol}}^{-1}}=1.000\text{mol.}[/latex]

[latex]\text{mol Al}=1\times 1.000\text{mol}=1.000\text{mol}[/latex]

Al_{2}Cl_{6}: [latex]\frac{266\text{g}}{266.6793\text{g}{\text{mol}}^{-1}}=0.997\text{mol}[/latex]

[latex]\text{mol Al}=2\times 0.997\text{mol}=1.994\text{mol}[/latex]

Al_{2}S_{3}: [latex]\frac{225\cancel{\text{g}}}{150.161\cancel{\text{g}}{\text{mol}}^{-1}}=1.50\text{mol}[/latex]

[latex]\text{mol Al}=2\times 1.50\text{mol}=3.00\text{mol}[/latex]

27. Determine the number of grams present in the diamond and from that the number of moles. Find the number of carbon atoms by multiplying Avogadro’s number by the number of moles:

[latex]\frac{3104\cancel{\text{carats}}\times \frac{200\cancel{\text{mg}}}{1\cancel{\text{carat}}}\times \frac{1\cancel{\text{g}}}{1000\cancel{\text{mg}}}}{12.011\cancel{\text{g}}\cancel{{\text{mol}}^{-1}}\left(6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\right)}=3.113\times {10}^{25}\text{C atoms}[/latex]

29. Determine the molar mass of sugar. 12(12.011) + 22(1.00794) + 11(15.9994) = 342.300 g/mol; Then [latex]0.0278\text{mol}\times 342.300\text{g/mol}=9.52\text{g sugar.}[/latex] This 9.52 g of sugar represents [latex]\frac{11.0}{60.0}[/latex] of one serving or [latex]\frac{60.0\text{g serving}}{11.0\cancel{\text{g sugar}}}\times 9.52\cancel{\text{g sugar}}=51.9\text{g cereal.}[/latex]

This amount is [latex]\frac{51.9\text{g cereal}}{60.0\text{g serving}}=0.865[/latex] servings, or about 1 serving.

31. Calculate the number of moles of each species, then remember that 1 mole of anything [latex]=6.022\times {10}^{23}[/latex] species. (a) 20.0 g = 1.11 mol H_{2}O; (b) 77.0 g CH_{4} = 4.79 mol CH_{4}; (c) 68.0 g CaH_{2} = 1.62 mol CaH_{2}; (d) 100.0 g N_{2}O = 2.27 mol N_{2}O; (e) 84.0 g HF = 4.20 mol HF. Therefore, 20.0 g H_{2}O represents the least number of molecules since it has the least number of moles.

### Glossary

**Avogadro’s number (**

*N*)_{A}experimentally determined value of the number of entities comprising 1 mole of substance, equal to [latex]6.022\times {10}^{23}{\text{mol}}^{-1}[/latex]

**formula mass**

sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass

**molar mass**

mass in grams of 1 mole of a substance

**mole**

amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of ^{12}C

- Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.”
*Scientific Report*3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447. ↵