# 4.5: Mass Ratio Calculation

One of the fundamental laws of chemistry deals with the fact that we cannot (using chemical means) create or destroy matter. When a reaction is run, the number of atoms of each specific type must be the same on both sides of the equation. For some materials, it turns out that one element can combine with a second element in more than one ratio. Carrying out mass ratio calculations helped establish the law of multiple proportions.

### Mass Ratio Calculations

Example 4.5.1

Copper reacts with chlorine to form two compounds. Compound A consists of $$4.08 \: \text{g}$$ of copper for every $$2.28 \: \text{g}$$ of chlorine. Compound B consists of $$7.53 \: \text{g}$$ of copper for every $$8.40 \: \text{g}$$ of chlorine. What is the lowest whole number mass ratio of copper that combines with a given mass of chlorine?

Solution:

Step 1: List the known quantities and plan the problem.

Known

• Compound A $$= 4.08 \: \text{g} \: \ce{Cu}$$ and $$2.28 \: \text{g} \: \ce{Cl}$$
• Compound B $$= 7.53 \: \text{g} \: \ce{Cu}$$ and $$8.40 \: \text{g} \: \ce{Cl}$$

Apply the law of multiple proportions to the two compounds. For each compound, find the grams of copper that combine with $$1.00 \: \text{g}$$ of chlorine by dividing the mass of copper by the mass of chlorine. Then find the ratio of the masses of copper in the two compounds by dividing the larger value by the smaller value.

Step 2: Calculate

$\text{Compound A} \: \frac{4.08 \: \text{g} \: \ce{Cu}}{2.28 \: \text{g} \: \ce{Cl}} = \frac{1.79 \: \text{g} \: \ce{Cu}}{1.00 \: \text{g} \: \ce{Cl}}$

$\text{Compound B} \: \frac{7.53 \: \text{g} \: \ce{Cu}}{8.40 \: \text{g} \: \ce{Cl}} = \frac{0.896 \: \text{g} \: \ce{Cu}}{1.00 \: \text{g} \: \ce{Cl}}$

Compare the masses of copper per gram of chlorine in the two samples.

$\frac{1.79 \: \text{g} \: \ce{Cu} \: \text{(in compound A)}}{0.896 \: \text{g} \: \ce{Cu} \: \text{(in compound B)}} = \frac{2.00}{1} = 2:1$

The mass ratio of copper per gram of chlorine in the two compounds is 2:1.

$$\ce{CuCl_2}$$.