READ: Molar Proportions

READ: Molar Proportions

Introduction

You have learned that chemical equations provide us with information about the types of particles that react to form products. Chemical equations also provide us with the relative number of particles and moles that react to form products. In this chapter, you will explore the quantitative relationships that exist between the reactants and products in a balanced equation. This is known as stoichiometry.

Stoichiometry involves calculating the quantities of reactants or products in a chemical reaction using the relationships found in the balanced chemical equation. The word stoichiometry actually comes from two Greek words: stoikheion, which means element, and metron, which means measure.

Interpreting Chemical Equations

Recall that a mole is a quantitative measure equivalent to Avogadro’s number of particles. How does the mole relate to the chemical equation? Consider the following reaction:

N2O3+H2O→2HNO3

We have learned that the coefficients in a chemical equation tell us the relative amounts of each substance involved in the reaction. One way to describe the ratios involved in the reaction above would be, “One molecule of dinitrogen trioxide, N₂O₃, plus one molecule of water yields two molecules of nitrous acid, HNO₃.” However, because these are only ratios, this statement would be equally valid using units other than molecules. As a result, we could also say, “One mole of dinitrogen trioxide plus one mole of water yields two moles of nitrous acid.”

We can use moles instead of molecules, because a mole is simply an amount equal to Avogadro’s number, just like a dozen is an amount equal to 12. It is important to not use units that describe properties other than amount. For example, it would notbe correct to say that one gram of dinitrogen trioxide plus one gram of water yields two grams of nitrous acid.

Now consider this reaction:

2CuSO4+4KI→2CuI+4K2SO4+I2

Here, we can say, “Two moles of copper (II) sulfate react with four moles of potassium iodide, yielding two moles of copper(I) iodide, four moles of potassium sulfate, and one mole of molecular iodine.” Although we can refer to molecules of iodine, I₂, it is generally not correct to refer to molecules of something like KI. Because KI is an ionic substance that exists as crystal lattices instead of discrete molecules, formula unit is used instead.

Using a Balanced Reaction to Compare Molar Quantities

A mole ratio is the relationship between two components of a chemical reaction. For instance, one way we could read the following reaction is that 2 moles of H_2(g)react with 1 mole of O_2(g) to produce 2 moles of H₂O_(l). 2H2(g)+O2(g)→2H2O(l)

The mole ratio of H_2(g) to O_2(g) would be: 

2 mol H21 mol O2

What is the ratio of hydrogen molecules to water molecules? By examining the balanced chemical equation, we can see that the coefficient in front of the hydrogen is 2, while the coefficient in front of water is also 2. Therefore, the mole ratio can be written as:

2 mol H2O1 mol O2

Example 1

Find the mole ratios for (1) calcium carbide to water and (2) calcium carbide to calcium hydroxide, given the balanced reaction:

CaC2(s)+2H2O(l)→Ca(OH)2(aq)+C2H2(g)

Solution: 

1. Mole ratio of calcium carbide to water =1 mol CaC22 mol H2O or 2 mol H2O1 mol CaC2 
2. Mole ratio of calcium carbide to calcium hydroxide =1 mol CaC21 mol Ca(OH)2 

The correct mole ratios of the reactants and products in a chemical equation are determined by the balanced equation. Therefore, the chemical equation must always be balanced before the mole ratios are used for calculations. 

Mole-Mole Calculations

In the chemistry lab, we rarely work with exactly one mole of a chemical. In order to determine the amount of reagent (reacting chemical) necessary or the amount of product expected for a given reaction, we need to do calculations using mole ratios.

Look at the following equation. If only 0.50 moles of magnesium hydroxide, Mg(OH)₂ are present, how many moles of phosphoric acid, H₃PO₄ would be required for the reaction?

2H3PO4+3Mg(OH)2→Mg3(PO4)2+6H2O

Step 1: To determine the conversion factor, we want to convert from moles of Mg(OH)₂ to moles of H₃PO₄. Therefore, the conversion factor is:

mole ratio=2 mol H3PO43 mol Mg(OH)2

Note that what we are trying to calculate is in the numerator, while what we know is in the denominator. 

Step 2: Use the conversion factor to answer the question.

Therefore, if we have 0.50 mol of Mg(OH)₂, we would need 0.33 mol of H₃PO₄ to react with all of the magnesium hydroxide. Notice if the equation was not balanced, the amount of H₃PO₄ required would have been calculated incorrectly. The ratio would have been 1:1, and we would have concluded that 0.50 mol of H₃PO₄ were required.

Example 2

How many moles of sodium oxide (Na₂O) can be formed from 2.36 mol of sodium nitrate (NaNO₃) using the balanced chemical equation below?

10Na+2NaNO3→6Na2O+N2O

Solution: (2.36 mol NaNO3).(6 mol Na2O2 mol NaNO3)=7.08 mol Na2O

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