# 8.4: Molar Ratios and Mole-to-Mole Conversions

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⚙️ Learning Objectives

• Use a balanced chemical equation to determine molar relationships between substances.

The previous section described the molar interpretation of a balanced chemical equation. The balanced chemical equation:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

tells us that 1 mol N2 reacts with 3 mol H2 to yield 2 mol NH3. Referring back to the automobile factory analogy, this also means that several relationships and possible conversions may be identified. When the coefficients from the balanced chemical equation are used to represent conversion factors in terms of the number of moles, the relationships are called molar ratios.

Relationships Molar Ratios

1 mol N2: 3 mol H2

$$\dfrac{1\;\mathrm{mol}\;{\mathrm N}_2}{3\;\mathrm{mol}\;{\mathrm H}_2}$$ $$\dfrac{3\;\mathrm{mol}\;{\mathrm H}_2}{1\;\mathrm{mol}\;{\mathrm N}_2}$$

1 mol N2: 2 mol NH3

$$\dfrac{1\;\mathrm{mol}\;{\mathrm N}_2}{2\;\mathrm{mol}\;{\mathrm{NH}}_3}$$ $$\dfrac{2\;\mathrm{mol}\;{\mathrm{NH}}_3}{1\;\mathrm{mol}\;{\mathrm N}_2}$$

3 mol H2: 2 mol NH3

$$\dfrac{3\;\mathrm{mol}\;{\mathrm H}_2}{2\;\mathrm{mol}\;{\mathrm{NH}}_3}$$ $$\dfrac{2\;\mathrm{mol}\;{\mathrm{NH}}_3}{3\;\mathrm{mol}\;{\mathrm H}_2}$$

The molar ratios may then be used to convert between moles of one substance and moles of a different substance, in much the same way that we were able to convert between wheels and cars, bumpers and wheels, or cars and bumpers.

Suppose we wanted to know how many moles of H2 were needed to completely react with 17.6 mol N2. The appropriate molar ratio is used in such a manner that units cancel when calculating the the moles of H2.

$$17.6\;\cancel{\mathrm{mol}\;{\mathrm N}_2}\times\dfrac{3\;\mathrm{mol}\;{\mathrm H}_2}{1\;\cancel{\mathrm{mol}\;{\mathrm N}_2}}=\boxed{52.8\;\mathrm{mol}\;{\mathrm H}_2}$$

Suppose we wanted to know how many moles of NH3 could be made when 0.773 mol H2 reacted with an excess of N2. Once again, the appropriate molar ratio is used in such a manner that units cancel when calculating the the moles of NH3. Note that the amount of N2 is irrelevant, since it is present in excess.

$$0.773\;\cancel{\mathrm{mol}\;{\mathrm H}_2}\times\dfrac{2\;\mathrm{mol}\;{\mathrm{NH}}_3}{3\;\cancel{\mathrm{mol}\;{\mathrm H}_2}}=\boxed{0.515\;\mathrm{mol}\;{\mathrm{NH}}_3}$$

✅ Example $$\PageIndex{1}$$

Identify all of the possible molar ratios for the balanced chemical equation:

2 H2 (g) + O2 (g) → 2 H2O (g)

Solution

Relationships Molar Ratios

2 mol H2: 1 mol O2

$$\dfrac{2\;\mathrm{mol}\;{\mathrm H}_2}{1\;\mathrm{mol}\;{\mathrm O}_2}$$ $$\dfrac{1\;\mathrm{mol}\;{\mathrm O}_2}{2\;\mathrm{mol}\;{\mathrm H}_2}$$

2 mol H2: 2 mol H2O

$$\dfrac{\cancel2\;\mathrm{mol}\;{\mathrm H}_2}{\cancel2\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}=\dfrac{\displaystyle1\;\mathrm{mol}\;{\mathrm H}_2}{\displaystyle1\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}$$ $$\dfrac{\cancel2\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}{\cancel2\;\mathrm{mol}\;{\mathrm H}_2}=\dfrac{\displaystyle1\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}{\displaystyle1\;\mathrm{mol}\;{\mathrm H}_2}$$

1 mol O2: 2 mol H2O

$$\dfrac{1\;\mathrm{mol}\;{\mathrm O}_2}{2\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}$$ $$\dfrac{2\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}{1\;\mathrm{mol}\;{\mathrm O}_2}$$

✅ Example $$\PageIndex{2}$$

Calculate the number of moles of hydrogen and oxygen required to make 3.34 mol of H2O according to the balanced chemical equation:

2 H2 (g) + O2 (g) → 2 H2O (g)

Solution

The appropriate molar ratios (see Example $$\PageIndex{1}$$) are used in such a manner that units cancel when calculating moles of H2 and moles of O2.

$$3.34\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}\times\dfrac{1\;\mathrm{mol}\;{\mathrm H}_2}{1\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}}=\boxed{3.34\;\mathrm{mol}\;{\mathrm H}_2}$$

$$3.34\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}\times\dfrac{1\;\mathrm{mol}\;{\mathrm O}_2}{2\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}}=\boxed{1.67\;\mathrm{mol}\;{\;\mathrm O}_2}$$

✏️ Exercise $$\PageIndex{1}$$

1. Write a balanced chemical equation that shows the reaction between aluminum metal and oxygen gas to form an aluminum oxide solid.
2. Identify all of the possible molar ratios.
4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
Relationships Molar Ratios

4 mol Al: 3 mol O2

$$\dfrac{4\;\mathrm{mol}\;\mathrm{Al}}{3\;\mathrm{mol}\;{\mathrm O}_2}$$ $$\dfrac{3\;\mathrm{mol}\;{\mathrm O}_2}{4\;\mathrm{mol}\;\mathrm{Al}}$$

4 mol Al: 2 mol Al2O3

$$\dfrac{\cancel4\;\mathrm{mol}\;\mathrm{Al}}{\cancel2\;\mathrm{mol}\;{\mathrm{Al}}_2{\mathrm O}_3}=\dfrac{2\;\mathrm{mol}\;\mathrm{Al}}{1\;\mathrm{mol}\;{\mathrm{Al}}_2{\mathrm O}_3}$$ $$\dfrac{\cancel2\;\mathrm{mol}\;{\mathrm{Al}}_2{\mathrm O}_3}{\cancel4\;\mathrm{mol}\;\mathrm{Al}}=\dfrac{1\;\mathrm{mol}\;{\mathrm{Al}}_2{\mathrm O}_3}{2\;\mathrm{mol}\;\mathrm{Al}}$$

3 mol O2: 2 mol Al2O3

$$\dfrac{3\;\mathrm{mol}\;{\mathrm O}_2}{2\;\mathrm{mol}\;{\mathrm{Al}}_2{\mathrm O}_3}$$ $$\dfrac{2\;\mathrm{mol}\;{\mathrm{Al}}_2{\mathrm O}_3}{3\;\mathrm{mol}\;{\mathrm O}_2}$$

✏️ Exercise $$\PageIndex{2}$$

Given the balanced chemical equation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

1. How many moles of O2 are needed to completely react with 3.08 mol C8H18?
2. How many moles of C8H18 will react when 7.7 mol CO2 are produced?
38.5 mol O2
0.96 mol C8H18

#### Summary

• The balanced chemical reaction can be used to determine molar relationships between substances.

This page is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Lance S. Lund (Anoka-Ramsey Community College), Marisa Alviar-Agnew, and Henry Agnew.

8.4: Molar Ratios and Mole-to-Mole Conversions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.