8.5: Mass-to-Mass Conversions

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⚙️ Learning Objectives

• Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.

Mole-to-Mass Conversions

We have established that a balanced chemical equation is balanced in terms of moles, as well as atoms or molecules. We have used balanced equations to set up ratios, in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions – such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that relation to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:

$${\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{moles}\;\mathrm{of}\;\mathrm A\;}}\;\xrightarrow[\mathrm{ratio}]{\mathrm{molar}}\;{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\mathrm{moles}\;\mathrm{of}\;\mathrm B\;}}\;\xrightarrow[\mathrm{mass}]{\mathrm{molar}}\;{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\mathrm{grams}\;\mathrm{of}\;\mathrm B\;}}$$

Collectively, these conversions are called mole-to-mass calculations. As an example, consider the balanced chemical equation

Fe2O3 + 3 SO3 → Fe2(SO4)3

Suppose we wanted to know how many grams of SO3 will react with 3.59 mol of Fe2O3. Using the mole-mass calculation sequence, we can determine the required mass of SO3 in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3 needed. From here, the mass of SO3 may be calculated using the molar mass of SO3 as a conversion factor.

Relationship Possible Conversions
molar ratio 1 mol Fe2O3: 3 mol SO3 $$\dfrac{1\;\mathrm{mol}\;{\mathrm{Fe}}_2{\mathrm O}_3}{3\;\mathrm{mol}\;{\mathrm{SO}}_3}$$ or $$\dfrac{3\;\mathrm{mol}\;{\mathrm{SO}}_3}{1\;\mathrm{mol}\;{\mathrm{Fe}}_2{\mathrm O}_3}$$
molar mass 1 mol SO3 = 80.07 g SO3 $$\dfrac{80.07\;\mathrm g\;{\mathrm{SO}}_3}{1\;\mathrm{mol}\;{\mathrm{SO}}_3}$$ or $$\dfrac{1\;\mathrm{mol}\;{\mathrm{SO}}_3}{80.07\;\mathrm g\;{\mathrm{SO}}_3}$$

$$3.59\:\cancel{\mathrm{mol}\:{\mathrm{Fe}}_2{\mathrm O}_3}\times\dfrac{3\:\mathrm{mol}\:{\mathrm{SO}}_3}{1\:\cancel{\mathrm{mol}\:{\mathrm{Fe}}_2{\mathrm O}_3}}=10.\underline77\:\mathrm{mol}\:{\mathrm{SO}}_3$$

Moles of Fe2O3 cancel, leaving moles of SO3 as the units. This answer may then be converted to grams of SO3:

$$10.\underline77\;\cancel{\mathrm{mol}\;{\mathrm{SO}}_3}\times\dfrac{80.07\:\mathrm g\:{\mathrm{SO}}_3}{1\;\cancel{\mathrm{mol}\;{\mathrm{SO}}_3}}=\boxed{862\:\mathrm g\:{\mathrm{SO}}_3}$$

The final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO3 will react with 3.59 mol of Fe2O3. Many problems of this type can be answered in this manner. It is generally easier if the problem is set up in a single series of steps:

$$3.59\cancel{\,\mathrm{mol}\,{\mathrm{Fe}}_2{\mathrm O}_3}\times\dfrac{3\;\cancel{\mathrm{mol}\;{\mathrm{SO}}_3}}{1\;\cancel{\mathrm{mol}\;{\mathrm{Fe}}_2{\mathrm O}_3}}\times\dfrac{80.07\;\mathrm g\;{\mathrm{SO}}_3}{1\;\cancel{\mathrm{mol}\;{\mathrm{SO}}_3}}=\boxed{862\,\mathrm g\,{\mathrm{SO}}_3}$$

Notice that the identical answer is obtained, in much the same manner as when multi-step dimensional analysis was introduced in Section 2.7.

✅ Example $$\PageIndex{1}$$: Generation of Aluminum Oxide

How many moles of HCl will be produced when 249 g of AlCl3 are reacted according to this chemical equation?

2 AlCl3 + 3 H2O → Al2O3 + 6 HCl

Solution

Steps for Problem Solving

Identify the "given" information and what the problem is asking you to "find."

Given: 249 g AlCl3
Find: moles HCl

List other known quantities.

1 mol AlCl3 = 133.33 g AlCl3
2 mol AlCl3: 6 mol HCl

Prepare a concept map using the proper conversion factor(s).

$${\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\;\mathrm g\;{\mathrm{AlCl}}_3\;\;}}\xrightarrow[{133.33\;\mathrm g\;{\mathrm{AlCl}}_3}]{1\;\mathrm{mol}\;{\mathrm{AlCl}}_3}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\mathrm{mol}\;{\mathrm{AlCl}}_3}}\xrightarrow[{2\;\mathrm{mol}\;{\mathrm{AlCl}}_3}]{6\;\mathrm{mol}\;\mathrm{HCl}}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\mathrm{mol}\;\mathrm{HCl}\;}}$$

Cancel units and calculate.

$$249\;\cancel{\mathrm g\;{\mathrm{AlCl}}_3}\times\dfrac{1\;\cancel{\mathrm{mol}\;{\mathrm{AlCl}}_3}}{133.33\;\,\cancel{\mathrm g\;{\mathrm{AlCl}}_3}}\times\dfrac{6\;\mathrm{mol}\,\mathrm{HCl}}{2\;\cancel{\mathrm{mol}\,{\mathrm{AlCl}}_3}}=\boxed{5.60\;\mathrm{mol}\;\mathrm{HCl}}$$

249 g AlCl3 is less than 266.66 g (the mass for 2 mol AlCl3).

Since the relationship is 6 mol HCl to 2 mol AlCl3, the answer should be less than 6 mol HCl.

The answer of 5.60 mol HCl is indeed less than 6 mol HCl.

✏️ Exercise $$\PageIndex{1}$$: Generation of Aluminum Oxide

How many moles of Al2O3 will be produced when 23.9 g of H2O are reacted according to this chemical equation?

2 AlCl3 + 3 H2O → Al2O3 + 6 HCl

0.442 mol Al2O3

Mass-to-Mass Conversions

In the real world, the number of molecules or moles are not measured directly. Rather, it is the masses of the reactants and/or products that will be measured directly. Consequently, it is much more practical to make all measurements in terms of the number of grams. It is a small step to move from mole-to-mass calculations to mass-to-mass calculations.

If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then – and only then – the balanced chemical equation is used to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:

$${\color[rgb]{0.5, 0.0, 0.5}\boxed{\;\mathrm{grams}\;\mathrm{of}\;\mathrm A\;}}\xrightarrow[\mathrm{mass}]{\mathrm{molar}}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{moles}\;\mathrm{of}\;\mathrm A\;}}\xrightarrow[\mathrm{ratio}]{\mathrm{molar}}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\mathrm{moles}\;\mathrm{of}\;\mathrm B\;}}\xrightarrow[\mathrm{mass}]{\mathrm{molar}}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\mathrm{grams}\;\mathrm{of}\;\mathrm B\;}}$$

This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. This text will generally approach this type of problem in a single series of steps.

✅ Example $$\PageIndex{2}$$: Decomposition of Ammonium Nitrate

Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.

NH4NO3 (s) → N2O (g) + 2 H2O (l)

Find the mass of each product formed when 45.7 g of ammonium nitrate is decomposed.

Solution

Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: 45.7 g NH4NO3

Find: Mass N2O, Mass H2O

List other known quantities.
1 mol NH4NO3 = 80.05 g NH4NO3

1 mol N2O = 44.02 g N2O

1 mol H2O = 18.02 g H2O

1 mol NH4NO3: 1 mol N2O

1 mol NH4NO3: 2 mol H2O

Prepare concept maps using the proper conversion factor(s).
$${\color[rgb]{0.5, 0.0, 0.5}\boxed{\;\;\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3\;\;}}\xrightarrow[{80.05\;\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3}]{1\;\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3\;}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}]{1\;\mathrm{mol}\;{\mathrm N}_2\mathrm O}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\mathrm{mol}\;{\mathrm N}_2\mathrm O\;\;\;}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm N}_2\mathrm O}]{44.02\;\mathrm g\;{\mathrm N}_2\mathrm O}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\;\;\;\mathrm g\;{\mathrm N}_2\mathrm O\;\;\;\;}}$$

$${\color[rgb]{0.5, 0.0, 0.5}\boxed{\;\;\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3\;\;}}\xrightarrow[{80.05\;\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3}]{1\;\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3\;}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}]{2\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\mathrm{mol}\;{\mathrm H}_2\mathrm O\;\;\;}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm H}_2\mathrm O}]{18.02\;\mathrm g\;{\mathrm H}_2\mathrm O}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\;\;\;\mathrm g\;{\mathrm H}_2\mathrm O\;\;\;\;}}$$

Cancel units and calculate.
$$45.7\:\cancel{\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3}\times\dfrac{1\;\cancel{\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}}{80.05\:\cancel{\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3}}\times\dfrac{1\;\cancel{\mathrm{mol}\;{\mathrm N}_2\mathrm O}}{1\:\cancel{\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}}\times\dfrac{44.02\;\mathrm g\;{\mathrm N}_2\mathrm O}{1\;\cancel{\mathrm{mol}\;{\mathrm N}_2\mathrm O}}=\boxed{25.1\;\mathrm g\;{\mathrm N}_2\mathrm O}$$

$$45.7\:\cancel{\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3}\times\dfrac{1\;\cancel{\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}}{80.05\:\cancel{\mathrm g\;{\mathrm{NH}}_4{\mathrm{NO}}_3}}\times\dfrac{2\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}}{1\:\cancel{\mathrm{mol}\;{\mathrm{NH}}_4{\mathrm{NO}}_3}}\times\dfrac{18.02\;\mathrm g\;{\mathrm H}_2\mathrm O}{1\;\cancel{\mathrm{mol}\;{\mathrm H}_2\mathrm O}}=\boxed{20.6\;\mathrm g\;{\mathrm H}_2\mathrm O}$$

The sum of the masses of the two products is equal to the mass of ammonium nitrate decomposed.

This demonstrates the Law of Conservation of Mass.

Each answer has three significant figures.

✏️ Exercise $$\PageIndex{2}$$: Carbon Tetrachloride

Methane reacts with elemental chlorine to make carbon tetrachloride, CCl4. The balanced chemical equation for this reaction is:

CH4 (g) + 4 Cl2 (g) → CCl4 (l) + 4 HCl (g)

How many grams of HCl are produced by the reaction of 100.0 g CH4?

909.1 g HCl

✅ Example $$\PageIndex{3}$$: Production of Ammonia

Taking into consideration the balanced chemical equation:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

what mass of hydrogen gas is needed to prepare 404 g of ammonia?

Solution

Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: 404 g NH3

Find: Mass H2

List other known quantities.
1 mol NH3 = 17.03 g NH3

1 mol H2 = 2.016 g H2

3 mol H2: 2 mol NH3

Prepare concept maps using the proper conversion factor(s).
$${\color[rgb]{0.5, 0.0, 0.5}\boxed{\;\;\mathrm g\;{\mathrm{NH}}_3\;\;}}\xrightarrow[{17.03\;\mathrm g\;{\mathrm{NH}}_3}]{1\;\mathrm{mol}\;{\mathrm{NH}}_3}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;{\mathrm{NH}}_3\;}}\xrightarrow[{2\;\mathrm{mol}\;{\mathrm{NH}}_3}]{3\;\mathrm{mol}\;{\mathrm H}_2}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\mathrm{mol}\;{\mathrm H}_2\;\;}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm H}_2}]{2.016\;\mathrm g\;{\mathrm H}_2}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\;\;\;\mathrm g\;{\mathrm H}_2\;\;\;\;}}$$

Cancel units and calculate.
$$404\:\cancel{\mathrm g\;{\mathrm{NH}}_3}\times\dfrac{1\;\cancel{\mathrm{mol}\;{\mathrm{NH}}_3}}{17.03\:\cancel{\mathrm g\;{\mathrm{NH}}_3}}\times\dfrac{3\;\cancel{\mathrm{mol}\;{\mathrm H}_2}}{2\:\cancel{\mathrm{mol}\;{\mathrm{NH}}_3}}\times\dfrac{2.016\;\mathrm g\;{\mathrm H}_2}{1\;\cancel{\mathrm{mol}\;{\mathrm H}_2}}=\boxed{71.7\;\mathrm g\;{\mathrm H}_2}$$

404 g NH3 is a bit more than 20 mol N3.

Since the molar ratio is 3 mol H2: 2 mol NH3, a bit more than 30 mol H2 are needed.

71.7 g H2 is a bit more than 30 mol H2.

✏️ Exercise $$\PageIndex{3}$$: Vanadium(V) Oxide

4 V (s) + 5 O2 (g) → 2 V2O5 (s)

1. What is the molar ratio between V and O2?
2. How many moles of V will react with 9.25 mol O2?
3. What mass of vanadium is needed to produce 0.301 g V2O5?
4 mol V: 5 mol O2
7.40 mol V
0.169 g V

Summary

• Calculations involving conversions between moles of a substance and the mass of that substance can be done using conversion factors.
• A balanced chemical reaction can be used to determine molar and mass 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.

8.5: Mass-to-Mass Conversions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.