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4.5: Mass Ratio Calculation

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     Two reactions involving the same reactants but in different proportions
    Figure \(\PageIndex{1}\) (Credit: Laura Guerin; Source: CK-12 Foundation; License: CC BY-NC 3.0(opens in new window))

    What are the similarities and differences between these two equations?

    One fundamental law of chemistry deals with the fact that we cannot create or destroy matter (using chemical means). 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 \(\PageIndex{1}\)

    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?

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

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

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

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

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

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

    Step 3: Think about your result.

    The ratio is a small whole-number ratio. For a given mass of chlorine, compound A contains twice the mass of copper as does compound B.

    Copper chloride crystals in a clear small glass from a top view.
    Figure \(\PageIndex{2}\): \(\ce{CuCl_2}\). (Credit: User:Chemicalinterest/Wikimedia Commons; Source: Commons Wikimedia, Copper Chloride Crystals (opens in new window) []; License: Public Domain)


    • The mass ratio gives the mass of an element that is found in combination with another element.


    1. What does the mass ratio between two elements tell us?
    2. If we compare the mass ratio of elements in one compound to that in an second compound what can we learn?  Give an example from the lesson above.
    3. In compound A, there is 6.3 g of hydrogen and 18.7 g of carbon, while in compound B there is 6.9 g of hydrogen and 41.0 g of carbon, what is the carbon to hydrogen mass ratio in each compound and how do these ratios compare?
    4. What are lowest ratios of hydrogen and carbon in compounds A and B? Predict their formulas

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