<meta http-equiv="refresh" content="1; url=/nojavascript/"> Stoichiometric Calculations | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Chemistry - Intermediate Go to the latest version.

12.2: Stoichiometric Calculations

Created by: CK-12

Lesson Objectives

  • Calculate the amount in moles of a reactant or product from the mass of another reactant or product. Calculate the mass of a reactant or product from the moles of another reactant or product.
  • Calculate the mass of a reactant or product from the mass of another reactant or product.
  • Create volume ratios from a balanced chemical equation.
  • Use volume ratios and other stoichiometric principles to solve problems involving mass, molar amounts, or volumes of gases.

Check Your Understanding

Recalling Prior Knowledge

  • How can a balanced chemical equation be used to construct mole ratios between substances?
  • How is a mole ratio used to convert from moles of one reactant or product to moles of another?

The mole ratio is the essence of ideal stoichiometry. The mole ratio tells us the quantitative relationship between reactants and products under ideal conditions, in which all reactants are completely converted into products. In the laboratory, most reactions are not completely ideal. Reactions may not proceed 100% to completion, or a given set of reactants may also undergo side reactions that lead to different products. However, theoretical stoichiometric calculations are important because they allow chemists to know the maximum possible amount of product that can be generated by a reaction from a given amount of each reactant.

Ideal Stoichiometry

Solving any stoichiometric calculation starts with a balanced chemical equation. As we saw in the last lesson, “Mole Ratios,” the coefficients of the balanced equation are the basis for the mole ratio between any pair of reactants and/or products. The following flowchart (Figure below) shows that the conversion from a given substance in moles to moles of an unknown substance involves multiplying by the relevant mole ratio.

This flowchart shows how a mole ratio is used in a stoichiometric conversion problem.

In this lesson, you will expand your understanding of stoichiometry to include the amounts of substances that are measured either by mass or by volume.

Mass-Based Stoichiometry

While the mole ratio is ever-present in all stoichiometry calculations, amounts of substances in the laboratory are most often measured by mass. Therefore, we need to use mole-mass calculations in combination with mole ratios to solve several different types of mass-based stoichiometry problems.

Mass to Moles Problems

In this type of problem, the mass of one substance is given, usually in grams. This value is then used to determine the amount in moles of another substance that will either react with or be produced from the given substance.

mass of given → moles of given → moles of unknown

The mass of the given substance is converted into moles by using the molar mass of that substance, which can be calculated from the atomic masses found on a periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation.

Sample Problem 12.3: Mass-Mole Stoichiometry

Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation:

Sn(s) + 2HF(g) → SnF2(s) + H2(g)

How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?

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

Known

  • given: 75.0 g Sn
  • molar mass of Sn = 118.69 g/mol
  • 1 mol Sn = 2 mol HF (mole ratio)

Unknown

  • mol HF

Use the molar mass of Sn to convert the given mass of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.

g Sn → mol Sn → mol HF

Step 2: Solve.

75.0 \ \text{g Sn} \times \dfrac{1 \ \text{mol Sn}}{118.69 \ \text{g Sn}} \times \dfrac{2 \ \text{mol HF}}{1 \ \text{mol Sn}} = 1.26 \ \text{mol HF}

Step 3: Think about your result.

The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.

Practice Problem
  1. Silver oxide is used in small batteries called button batteries (Figure below). It decomposes upon heating to form silver metal and oxygen gas.
2Ag2O(s) → 4Ag(s) + O2(g)

How many moles of silver are produced by the decomposition of 4.85 g of Ag2O?

Silver oxide is used in button batteries such as these watch batteries.

Moles to Mass Problems

In this type of problem, the amount of one substance is given in moles. From this, you can determine the mass of another substance that will either react with or be produced from the given substance.

moles of given → moles of unknown → mass of unknown

The moles of the given substance are first converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Then, the moles of the unknown are converted into a mass (in grams) by using its molar mass, which again can be calculated from the atomic masses given on the periodic table.

Sample Problem 12.4: Mole-Mass Stoichiometry

Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor.

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)

What mass of oxygen gas is consumed in a reaction that produces 4.60 moles of SO2?

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

Known

  • given: 4.60 mol SO2
  • 2 mol SO2 = 3 mol O2 (mole ratio)
  • molar mass of O2 = 32.00 g/mol

Unknown

  • mass O2 = ? g

Use the mole ratio to convert from mol SO2 to mol O2. Then convert mol O2 to grams. This will be done in a single two-step calculation.

mol SO2 → mol O2 → g O2

Step 2: Solve.

4.60 \ \text{mol SO}_2 \times \dfrac{3 \ \text{mol O}_2}{2 \ \text{mol SO}_2} \times \dfrac{32.00 \ \text{g O}_2}{1 \ \text{mol O}_2} = 221 \ \text{g O}_2

Step 3: Think about your result.

According to the mole ratio, 6.90 mol O2 is produced, which has a mass of 221 g. The answer has three significant figures because the given value has three significant figures.

Practice Problem
  1. Copper metal reacts with sulfur to form copper(I) sulfide. What mass of copper(I) sulfide is produced by the reaction of 0.528 mol Cu with excess sulfur?

Mass to Mass Problems

Mass-mass calculations are the most practical of all mass-based stoichiometry problems. Moles cannot be measured directly, but masses can be easily measured in the lab for most substances. This type of problem is three steps and is a combination of the two previous types.

mass of given → moles of given → moles of unknown → mass of unknown

The mass of the given substance is converted into moles by using its molar mass. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Finally, the moles of the unknown are converted back to a mass by using its molar mass.

Sample Problem 12.5: Mass-Mass Stoichiometry

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

NH4NO3(s) → N2O(g) + 2H2O(l)

In a certain experiment, 45.7 g of ammonium nitrate is decomposed. Find the mass of each of the products formed.

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

Known

  • given: 45.7 g NH4NO3
  • 1 mol NH4NO3 = 1 mol N2O = 2 mol H2O (mole ratios)
  • molar mass of NH4NO3 = 80.06 g/mol
  • molar mass of N2O = 44.02 g/mol
  • molar mass of H2O = 18.02 g/mol

Unknown

  • mass N2O = ? g
  • mass H2O = ? g

Perform two separate three-step mass-mass calculations, as shown below.

g NH4NO3 → mol NH4NO3 → mol N2O → g N2O
g NH4NO3 → mol NH4NO3 → mol H2O → g H2O

Step 2: Solve.

45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{1 \ \text{mol N}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{44.02 \ \text{g N}_2\text{O}}{1 \ \text{mol N}_2\text{O}} = 25.1 \ \text{g N}_2\text{O}
45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{2 \ \text{mol H}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{18.02 \ \text{g H}_2\text{O}}{1 \ \text{mol H}_2\text{O}} = 20.6 \ \text{g H}_2\text{O}

Step 3: Think about your result.

The total mass of the two products is equal to the mass of ammonium nitrate that decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures.

Practice Problem
  1. Solid iron(III) hydroxide reacts with sulfuric acid to produce aqueous iron(III) sulfate and water. What mass of sulfuric acid is needed to completely react with 12.72 g of iron(III) hydroxide? What mass of iron(III) sulfate is produced?

Volume-Based Stoichiometry

As you learned in the chapter on The Mole, Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain the same number of gas particles. We also saw that one mole of any gas at standard temperature and pressure (0°C and 1 atm) occupies a volume of 22.4 L. These characteristics make stoichiometry calculations involving gases at STP very straightforward. Consider the following reaction, in which nitrogen and oxygen combine to form nitrogen dioxide.

& \text{N}_{2}(g) && + && 2\text{O}_{2}(g) && \rightarrow && 2\text{NO}_{2}(g) \\& 1 \ \text{molecule} && && 2 \ \text{molecules} && && 2 \ \text{molecules} \\& 1 \ \text{mol} && && 2 \ \text{mol} && && 2 \ \text{mol} \\& 1 \ \text{volume} && && 2 \ \text{volumes} && && 2 \ \text{volumes}

Because of Avogadro’s hypothesis, we know that mole ratios between substances in a gas-phase reaction are also volume ratios. The six possible volume ratios for the above equation are:

  • \dfrac{1 \ \text{volume N}_2}{2 \ \text{volumes O}_2} or \dfrac{2 \ \text{volumes O}_2}{1 \ \text{volume N}_2}
  • \dfrac{1 \ \text{volume N}_2}{2 \ \text{volumes NO}_2} or \dfrac{2 \ \text{volumes NO}_2}{1 \ \text{volume N}_2}
  • \dfrac{2 \ \text{volumes O}_2}{2 \ \text{volumes NO}_2} or \dfrac{2 \ \text{volumes NO}_2}{2 \ \text{volumes O}_2}

Volume to Volume Problems

The volume ratios above can easily be used when the volume of one gas in a reaction is known and you need to determine the volume of another gas that will either react with or be produced from the first gas. Although pressure and temperature need to be held constant over the course of the reaction for these conversion factors to remain true, the reaction does not need to be run at STP; Avogadro's hypothesis is true regardless of the pressure and temperature being used.

Sample Problem 12.6: Volume-Volume Stoichiometry

The combustion of propane gas produces carbon dioxide and water vapor (see Figure below).

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

A municipal propane tank in Austin, TX. The combustion of propane gas produces carbon dioxide and water vapor.

What volume of oxygen is required to completely combust 0.650 L of propane? What volume of carbon dioxide is produced in the reaction?

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

Known

  • given: 0.650 L C3H8
  • 1 volume C3H8 = 5 volumes O2
  • 1 volume C3H8 = 3 volumes CO2

Unknown

  • volume O2 = ? L
  • volume CO2 = ? L

Step 2: Solve.

0.650 \ \text{L C}_3\text{H}_8 \times \frac{5 \ \text{L O}_2}{1 \ \text{L C}_3\text{H}_8} = 3.25 \ \text{L O}_2
0.650 \ \text{L C}_3\text{H}_8 \times \frac{3 \ \text{L CO}_2}{1 \ \text{L C}_3\text{H}_8} = 1.95 \ \text{L CO}_2

Step 3: Think about your result.

Because the coefficients on O2 and CO2 are larger than the one in front of C3H8, the volumes for those two gases are greater. Note that total volume is not necessarily conserved in a reaction because moles are not necessarily conserved. In this reaction, 6 total volumes of reactants become 7 total volumes of products.

Practice Problem
  1. Using the equation for the combustion of propane from Sample Problem 12.6, what volume of water vapor is produced in a reaction that generates 320. mL of carbon dioxide?

Mass to Volume and Volume to Mass Problems

Chemical reactions frequently involve both solid substances whose mass can be measured as well as gases for which measuring the volume is more appropriate. Stoichiometry problems of this type are called either mass-volume or volume-mass problems.

mass of given → moles of given → moles of unknown → volume of unknown
volume of given → moles of given → moles of unknown → mass of unknown

Because both types of problems involve a conversion from either moles of gas to volume or vice-versa, we can use the molar volume of 22.4 L/mol as a conversion factor if the reaction is run at STP. In a later chapter (The Behavior of Gases), we will see how to solve this type of problem when other sets of reaction conditions are used.

Sample Problem 12.7: Mass-Volume Stoichiometry

Aluminum metal reacts rapidly with aqueous sulfuric acid to produce aqueous aluminum sulfate and hydrogen gas.

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

Determine the volume of hydrogen gas produced at STP when a 2.00 g piece of aluminum completely reacts with sulfuric acid.

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

Known

  • given: 2.00 g Al
  • molar mass of Al = 26.98 g/mol
  • 2 mol Al = 3 mol H2

Unknown

  • volume H2 = ?

The grams of aluminum will first be converted to moles. Then the mole ratio will be applied to convert to moles of hydrogen gas. Finally, the molar volume of a gas will be used to convert to liters of hydrogen.

g Al → mol Al → mol H2 → L H2

Step 2: Solve.

2.00 \ \text{g Al} \times \frac{1 \ \text{mol Al}}{26.98 \ \text{g Al}} \times \frac{3 \ \text{mol H}_2}{2 \ \text{mol Al}} \times \frac{22.4 \ \text{L H}_2}{1 \ \text{mol H}_2} = 2.49 \ \text{L H}_2

Step 3: Think about your result.

The volume result is in liters. For smaller amounts, it may be convenient to convert to milliliters. The answer here has three significant figures. Because the molar volume is a measured quantity of 22.4 L/mol, three is the maximum number of significant figures for this type of problem.

Sample Problem 12.8: Volume-Mass Stoichiometry

Calcium oxide is used to trap the sulfur dioxide that is generated in coal-burning power plants according to the following reaction:

2CaO(s) + 2SO2(g) + O2(g) → 2CaSO4(s)

What mass of calcium oxide is required to react completely with 1.2 × 103 L of sulfur dioxide at STP?

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

Known

  • given: 1.2 × 103 L SO2
  • 2 mol SO2 = 2 mol CaO
  • molar mass of CaO = 56.08 g/mol

Unknown

  • mass CaO = ? g

The volume of SO2 will first be converted to moles. The mole ratio can then be used, and finally, moles of CaO will be converted to grams.

L SO2 → mol SO2 → mol CaO → g CaO

Step 2: Solve.

1.2 \times 10^3 \ \text{L SO}_2 \times \frac{1 \ \text{mol SO}_2}{22.4 \ \text{L SO}_2} \times \frac{2 \ \text{mol CaO}}{2 \ \text{mol SO}_2} \times \frac{56.08 \ \text{g CaO}}{1 \ \text{mol CaO}} = 3.0 \times 10^3 \ \text{g CaO}

Step 3: Think about your result.

The resultant mass could also be reported as 3.0 kg, with two significant figures. Even though the 2:2 mole ratio does not mathematically affect the problem, it is still necessary for unit conversion.

Practice Problem
  1. Sodium azide (NaN3) is a compound that is used in automobile air bags (Figure below). A collision triggers its rapid decomposition into sodium and nitrogen gas, which is the gas that fills the air bag.
    1. The decomposition of 1.00 g of NaN3 produces what volume of N2 at STP?
    2. What mass of NaN3 is required to produce 250. L of N2 at STP?

The rapid decomposition of sodium azide produces nitrogen gas, which fills an automotive air bag.

Other Stoichiometry

Stoichiometric conversions all involve mole ratios between substances in a balanced chemical equation. Problems that involve mass and/or the volume of a gas are very common and practical. However, a third “arm” of the mole road map could also be part of a problem – the number of representative particles of a substance. Using the equation in Sample Problem 12.7, we could determine the number of formula units of aluminum sulfate produced when 25.0 g of Al reacts.

25.0 \ \text{g Al} \times \frac{1 \ \text{mol Al}}{26.98 \ \text{g Al}} \times \frac{1 \ \text{mol Al}_2(\text{SO}_4)_3}{2 \ \text{mol Al}} \times \frac{6.02 \times 10^{23} \ \text{form. units Al}_2(\text{SO}_4)_3}{1 \ \text{mol Al}_2(\text{SO}_4)_3} = 2.79 \times 10^{23} \ \text{form. units Al}_2(\text{SO}_4)_3

Problems could potentially arise involving any combination of mass, volume, and number of representative particles. However, since particles of this size cannot actually be counted and stoichiometry is used most often for lab-based situations, problems involving the number of particles are seldom encountered in the real world.

Summary of Stoichiometry

The flowchart in Figure below illustrates the types of stoichiometry problems that we have seen in this chapter and that you will most often need to solve. Conversion (b) is always present in any stoichiometry problem, while the use of the conversions represented by (a), (c), (d), and (e) depend on the specific type of problem. Conversion (f) is unique to gaseous volume-volume problems in which the pressure and temperature are held constant.

Flowchart for solving many types of stoichiometry problems.

To learn more about stoichiometric calculations, watch the video lecture at http://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiometry/v/stoichiometry.

An example of how to solve a stoichiometry problem can be seen at http://www.khanacademy.org/science/physics/thermodynamics/v/stoichiometry-example-problem-1.

A second example of how to solve a stoichiometry problem can be seen at http://www.khanacademy.org/science/physics/thermodynamics/v/stoichiometry-example-problem-2.

Lesson Summary

  • All stoichiometry problems involve the use of a mole ratio to convert between moles of a given substance and moles of an unknown substance.
  • In an ideal stoichiometry problem, the mass of any reactant or product can be calculated from the mass of any other reactant or product if the balanced equation is known.
  • Since the molar volume of any gas at STP is constant, gas volumes can also be used in stoichiometric calculations. This allows for multiple types of problems where amounts can be indicated by mass, volume, or number of moles.

Lesson Review Questions

Reviewing Concepts

  1. How many conversion factors are involved in each of the following stoichiometry problems, where A and B are two components of a chemical reaction, and all volumes are for gases at STP?
    1. moles of A → mass of B
    2. volume of A → volume of B
    3. mass of B → mass of A
    4. moles of B → volume of A
  2. Why does a mass-mass stoichiometry problem require three steps, while a volume-volume stoichiometry problem only requires one step?

Problems

  1. The double-replacement reaction that occurs when aqueous solutions of calcium chloride and silver nitrate are combined produces a solution of calcium nitrate and a precipitate of silver chloride.
    1. How many moles of calcium chloride are needed to react completely with 2.28 moles of silver nitrate?
    2. If 0.0623 moles of CaCl2 reacts completely, how many grams of AgCl are produced?
    3. In order to produce exactly 1.50 g of AgCl, how many moles of each of the two reactants should be used?
  2. Silicon dioxide reacts with carbon upon heating to produce silicon carbide (SiC) and carbon monoxide.
    1. What mass of carbon is required to react completely with 15.70 g of SiO2?
    2. When 152 g of SiO2 reacts with excess carbon, what mass of SiC is produced?
    3. If 42.2 g of CO were produced by this reaction, what mass of carbon must have reacted?
  3. Butane (C4H10) combusts according to the following reaction: Assume no change in temperature or pressure for the following questions.
    1. What volume of O2 is needed to combust 425 mL of butane?
    2. What volume of butane must be combusted to produce 729 L of CO2?
    3. When 6.20 L of butane is combusted, what volumes of CO2 and H2O are produced?
  4. Dissolving calcium carbonate in hydrochloric acid produces aqueous calcium chloride, carbon dioxide, and water. Assume the reaction takes place at STP.
    1. What volume of CO2 is produced by the reaction of 9.58 g CaCO3?
    2. If 67.1 L of CO2 is produced by this reaction, what mass of HCl must have reacted?
    3. If 0.812 mol of HCl is used in this reaction, what mass of CaCO3 would be consumed? What volume of CO2 would be produced?
  5. Nitroglycerin is an explosive compound that decomposes into multiple gaseous products according to the following reaction:
    1. What mass of nitrogen gas at STP is produced when 0.314 g of nitroglycerin decomposes?
    2. What volume of carbon dioxide at STP is produced in a reaction that also produces 2.25 moles of O2?
    3. What is the total volume of all gases produced at STP by the full decomposition of 10.0 g of nitroglycerin?
  6. Iron rusts to form iron(III) oxide according to the following equation:
    1. If 46.2 g of Fe2O3 are produced by this reaction, how many Fe atoms reacted?
    2. What volume of O2 at STP is needed to fully react with 8.39 × 1024 atoms of iron?
    3. The complete reaction of 0.916 mol Fe will produce how many formula units of Fe2O3?
  7. Zinc reacts with hydrochloric acid according to the following equation: In a certain experiment, a 3.77 g sample of impure zinc is reacted with excess hydrochloric acid. If 1.09 L of H2 gas is collected at STP, what percentage of the original sample was zinc? Assume that the impurities do not react with HCl.

Further Reading / Supplemental Links

Points to Consider

This lesson dealt with ideal stoichiometry, where 100% of the reactants were converted to products. In the real world, many chemical reactions do not proceed entirely in this way.

  • How can we calculate the amount of products that could be formed in a reaction when two or more reactants are combined in a ratio other than the mole ratio from the balanced equation?
  • How can we express the extent to which a set of reactants is converted to products if it is less than 100%?

Image Attributions

Files can only be attached to the latest version of None

Reviews

Please wait...
You need to be signed in to perform this action. Please sign-in and try again.
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.SCI.ENG.SE.1.Chemistry-Intermediate.12.2

Original text