### Learning Objectives

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

- Convert between mass and moles using stoichiometry
- Calculate the percent composition of a compound using experimental data.
- Calculate percent error.

**Introduction**

Many metals can react with acid to form hydrogen gas and a metal chloride. This reaction is given by:

**Equation 1**

[latex]\text{M}+\text{nHCl}\to\text{MCln}+\frac{n}{2}\text{H}_{2\left(g\right)}\\[/latex]

Where M is the metal used and n is the charge on the metal cation. For example, since Ca is an alkaline earth metal, we know it has a charge of +2 when it exists as a cation. Therefore the reaction between calcium and hydrochloric acid would be given as:

[latex]\text{Ca}+2\text{HCl}\to\text{CaCl}_2+\text{H}_{2\left(g\right)}\\[/latex]

The chemical equation for a reaction will give the mol ratios as well as the formulas for all species involved. We can use stoichiometry to convert between the mass and mol of each species in a dimensional analysis calculation.

The above reaction produces hydrogen gas and a metal chloride. The relative atomic mass of hydrogen is 1.008 g mol-1 which means the molar mass of hydrogen gas is ([latex]2\times1.008=2.016[/latex] g/mol). If we measure the amount of hydrogen gas produced by a known mass of a metal then, using the stoichiometry of the above reaction, we can calculate the relative atomic mass of the metal used in the reaction.

In this experiment we will carry out the reaction using one of three metals; aluminium, calcium, or magnesium. Since it would be difficult to measure the mass of hydrogen gas in lab, you will instead collect the hydrogen gas and record the volume. We can convert the volume of hydrogen gas to the mol using an equation. The following equation relates mol of a gas to its volume.

**Equation 2**

[latex]PV=nRT[/latex]

where P is the pressure of the laboratory in atmospheres, V is the volume of hydrogen gas collected in liters, n is mol of the hydrogen gas, and T is the temperature in Kelvin. R is a constant that has the value 0.08206 . You will obtain the pressure in the lab from the barometer. The volume will be measured using the graduated cylinder collecting the hydrogen gas. The temperature will be collected from your lab station. Once these numbers are collected (and converted to the appropriate units, the equation can be rearranged to solve for the mol of hydrogen gas produced).

**Equation 3**

[latex]\displaystyle{n}=\frac{\text{PV}}{\text{RT}}\\[/latex]

Using stoichiometry, we can calculate the moles of metal initially used in the reaction. This is the relationship given by the balanced equation. Remember the mol ratios of all substances are given by the coefficient in a balanced equation. Additionally, since the grams of metal are measured at the beginning of the experiment, it is possible to calculate the atomic mass by:

**Equation 4**

[latex]\displaystyle\frac{\text{grams}}{\text{mol}}=\text{Molar Mass}\\[/latex]

Another way to look at the chemical composition of a compound is to evaluate the percent composition of each element. The percent composition is represented below and is the grams of one element divided by the molar mass of the compound multiplied by 100.

**Equation 5**

[latex]\displaystyle\text{Percent Composition}=\frac{\text{grams element}}{\text{molar mass}}\times100\\[/latex]

For example sodium oxide has the formula [latex]{Na}_{0}\\[/latex]O. Calculating the molar mass we find that:

Atom | # | Mass | Total |

Na | 2 | 22.99 | 45.98 g Na |

O | 1 | 16.00 | 16.00 g O |

61.98 g/mol |

There are 61.98 grams of this compound in one mol. However only 45.98 grams of that molar mass is due to Na. We can therefore calculate the percent composition of the compound as:

[latex]\displaystyle\text{Percent Composition O}=\frac{16.00\text{ g}}{61.98}\times100=25.81\%\text{ O}\\[/latex]

[latex]\displaystyle\text{Percent Composition Na}=\frac{45.98\text{ g}}{61.98}\times100=74.18\%\text{ Na}\\[/latex]

This information gives insight to the chemical composition of the compound and can therefore be a useful quality to investigate.

In today’s experiment you will collect information and calculate the molar mass of the metal you used (using the equation and calculations mentioned above). The molar mass may not correspond exactly with the value for that element on the periodic table. For example, perhaps when you perform the experiment the molar mass of the metal is calculated as 36.55 g/mol. If you know the options for the metal are Al, Ca and Mg, you can compare this value you obtained theoretically to the values on the periodic table for those elements. This value is closest to the molar mass of Ca of 40.08.

It is not enough for chemists to know there is a difference between the value you should have obtained (the true value) and the value you calculated in lab (the experimental value). Instead, scientists want to look at the relative amount of the difference. Consider a scale that reads values of 15 lbs heavier than the true measurement. If you used this scale to measure your own body weight that 15 lbs would be a significant difference! If the scale was used to measure the weight of moving trucks (~ 7500 lbs) the difference would be relatively insignificant. In order to put the relative difference into perspective, it is necessary to calculate percent error. Percent error is given by equation 6:

**Equation 6**

[latex]\displaystyle\text{Percent Error}=\frac{\left|\text{Experimental Value}-\text{Theoretical Value}\right|}{\text{Theoretical Value}}\times100\\[/latex]

Where the absolute value of the difference between the experimental value and the true value is divided by the true value and multiplied by 100. Most of the time, it is not as important to note whether the error gives a value too large or too small compared to the theoretical value. It is most important to consider the relative (percent) size of the error obtained which is why the absolute value is taken of the top part of the equation. By this equation, the difference in the above experiment would give students a percent error of

[latex]\displaystyle\text{Percent Error}=\frac{\left|36.55-40.08\right|}{40.08}\times100-8.81\%\\[/latex]

This experiment will relate data obtained in lab to the concepts of stoichiometry, percent composition and percent error.