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Chemistry LibreTexts

Stoichiometry - A Review

Skills Tested by This Quiz

  • evaluate molecular weight for a given formula
  • evaluate weight (mass) percentages of elements for a given formula
  • evaluate amounts (in mass and mole units) produced in a chemical reaction from given conditions
  • classify reactions by types: combination, combustion, displacement, formation, etc
  • determine the chemical formula when weight percentages are given and then evaluate the mole percentages of elements in the formula
  • determine the chemical formula when weight percentages are given and molecular weight is known
  • determine the amount produced, the actual yield, and other stoichiometry quantities for a given reaction

Review Purposes

  • To get an overall view of stoichiometry.
  • Apply skills learned to perform quantitative chemical analysis.
  • Apply theories and rules of chemistry to solve problems.
  • Assess areas of strength and weakness for review purposes.
  • Improve problem solving strategy and learning efficiency.


STOICHIOMETRY is the quantitative relationship of reactants and products. This unit has been divided into the following objects. A brief review is given here so that you can get a birds'-eye or overall view of stoichiometry.

  1. Amounts of substances
    Express amounts of substance in mass units of g, kg, tons, and convert them to moles, kilomoles, or millimoles.
  2. Chemical formulas
    Represent a substance with a formula that reflects its chemical composition, structure, and bonding; evaluate weight and mole percentages of elements in a substance; and determine chemical formula by elemental analysis.
  3. Reaction features
    Define some common features of chemical reactions; classify chemical reactions by common features such as combination, combustion, decomposition, displacement, and redox reactions.
  4. Reaction equations
    Express quantitative relationships using chemical reaction equations; evaluate quantities of reactants and products in a chemical reaction; and solve reaction stoichiometry problems.
  5. Excess and limiting reagents
    Define excess and limiting reagents; determine excess and limiting reagents in a reaction mixture; and determine quantities produced in a chemical reaction.
  6. Yields
    Define theoretical and actual yields due to limiting reagent; apply the concept of limiting reagent to evaluate theoretical yield; convert actual yield to percentage yields.

A Typical Quiz

In these quizzes, we give a set of question that are interrelated.

Problem 1

You have weighed x g of \(\ce{CaCl2}\) (molar mass 111) for an experiment dealing with chloride ions and you need to find the number of moles of \(\ce{Cl-}\) ions. Show the conversion from x g \(\ce{CaCl2}\) to moles of \(\ce{Cl-}\) ions.

The conversion logic in given below:

\(\ce{x\: g\: CaCl2 \times \dfrac{1 mol\: CaCl2}{111 g\: CaCl2}\times\dfrac{2 mol\: Cl- }{1 mol\: CaCl2}=\dfrac{2 x}{111}mol\: Cl-}\)


Convert amounts of substances in various units.

Problem 2

You have a sample weighing 2.345 g containing equal weight of \(\ce{CaCl2}\) (f.wt. 111) and \(\ce{AlCl3}\) (f.wt. 133). How many mole of \(\ce{Cl-}\) ions do you have?

Assuming you understand the previous example, then you can give

\(\mathrm{\left(\dfrac{2\: mol}{111\: g}+\dfrac{3\: mol}{133\: g} \right )\times\dfrac{2.345\: g}{2}= 0.04757\: mol\: Cl^-}\)


Develop skills to calculate moles of chloride ions in problems containing unequal weights of \(\ce{CaCl2}\) and \(\ce{AlCl3}\).

Problem 3

When a 2.00-g mixture of \(\ce{CaCl2}\) (111) and \(\ce{RbCl}\) (121) was analyzed, the result reports that the mixture contains a total of 0.0241 mol \(\ce{Cl-}\) ions. Find weight percentage of \(\ce{RbCl}\) in the sample.

Using the method in the previous examples, and assuming x g \(\ce{RbCl}\) are present in the mixture, you have

\(\dfrac{x}{121}+\dfrac{2 (2.00-x)}{111}= 0.0241\)

Solving for x, x = 1.23 g (Practice your solution)

\(\mathrm{Percentage = \dfrac{1.23}{2.00} = 61.4 \%\: RbCl}\)

Making an assumption to make the problem solvable.

Problem 4

When a 2.00-g mixture of \(\ce{CaCl2}\) (111) and \(\ce{RbCl}\) (121) was received by an analytical laboratory, the technician dissolved it in a solution and then he used an excess amount of \(\ce{AgNO3}\) to precipitate the chloride as \(\ce{AgCl}\). He collected 3.456 g \(\ce{AgCl}\). What is the percentage of \(\ce{RbCl}\) in the sample?

We should convert 3.456 g \(\ce{AgCl}\) into moles of \(\ce{AgCl}\) or \(\ce{Cl-}\) ions.

\(\mathrm{3.456\: g\: AgCl\times\dfrac{1\: mol\: AgCl}{143.4\: g\: AgCl}= 0.0241\: mol\: AgCl\: or\: Cl^-}\)

The rest of the solution has already been given in Problem 3.

Building skills to do useful work.

Problem 5

You are interviewed for a manager's job, and you were asked what you should instruct your technician to do for the following situation. Company X has received a sample containing a mixture of \(\ce{CaCl2}\) and \(\ce{BaCl2}\). The client of Company X needs to know the percentages of the two components.

Instruct your technician to find out if the sample contains any other substance than \(\ce{CaCl2}\) and \(\ce{BaCl2}\).

If it does not contain other substances than \(\ce{Ca^2+}\), \(\ce{Ba^2+}\), and \(\ce{Cl-}\) ions, instruct him to precipitate the chloride ions with an excess amount of silver ion.

Instruct the technician how to calculate the moles of chloride present, and how to calculate the percentages.

Write instructions for laboratory process.

Example Problems

  1. When properly used, urea, \(\ce{CH4N2O}\), is a good candidate for fertilizer. Unocal is one of the companies that produces urea in Kenai, Alaska. It produces 1.1 million tons of urea each year. How many moles are this amount?

    Hint: 18 trillion moles

    Urea is also called carbamide or carbonyldiamide. Since this is a US factory, 1 ton is taken as 983 kg that is equivalent to 18 thousand moles.

    Skill -
    Converting from mass units to moles and vice versa.

  2. Urea is made by reacting carbon dioxide (\(\ce{CO2}\)) with anhydrous ammonia (\(\ce{NH3}\)) under 3,000 psi (pounds per square inch) pressure and at 350°F (177°C or 450 K).

    &CO2 + &&2 NH3 &&=\: &&CO(NH2)2 + &&H2O\\
    &\:44 &&\:\:34  &&= &&\:\:\:\:\:\:60 &&\:\:18 \:(molar\: mass)}

    The removal of water that occurs during the reaction is referred to as "dehydration." The resulting molten mixture is further processed into either prills or granules. For each ton of urea made, how much water is removed?

    Hint: 0.3 tons

    For 1.3 tons of reactants, 1.0 ton of urea is produced. By the way, 3,000 psi is equivalent to 200 atmosphere pressure.

    Skill -
    Interpret stoichiometry among products and reactants.

  3. The NPK numbers (for example, 25, 3, 7) used in the fertilizer industry refer to the percentage of nitrogen, \(\ce{N}\), phosphorus, \(\ce{P}\), and potassium, \(\ce{K}\). What are the \(\ce{NPK}\) numbers for pure urea?

    Hint: 46 0 0

    Based on the formula, urea contains 46.7% \(\ce{N}\). However, the factory usually gives 46 because most products contain water.

    Skill -
    Calculate weight and mole percentage from chemical formula.

  4. Conversion of nitrogen into protein by plants is a complicated process. First, urea is hydrolyzed and converted to ammonium carbonate. The hydrolysis of urea is brought about by the urease enzyme, but the reaction may be written as follows:

    \(\ce{CO(NH2)2 + 2 H2O \rightarrow (NH4)2CO3}\)

    The product is a salt. The solubility of urea is reported to be 1 g in 1 ml of water. Assuming that this solution is allowed to hydrolyze by the urease enzyme, which is the limiting reagent, urea or water?

    Hint: urea

    One g of urea is 17 millimoles, which requires 34 millimoles of water to react from the stoichiometry of the reaction equation. One g of water is 56 millimoles; that is more than 34 millimoles.

    Skill -
    Determine excess and limiting reagent.

  5. Alanine \(\ce{CH3CH(NH2)(COOH)}\) is one of the amino acids that are the building blocks of proteins. What is the percentage of nitrogen in this amino acid?

    Hint: 16% \(\ce{N}\) in alanine

    Skill -
    Determine an element percentage from chemical formula.

  6. An old method for protein analysis was to measure the nitrogen content. The amount of protein present was assumed to be 16 times the amount of nitrogen.

    Assume that 50 % of the nitrogen in urea is converted to protein by plants. How much protein is produced for one gram of urea?

    Hint: 3.68 g protein

    Percentages of protein in plant material are usually very low.

    Skill -
    Calculate theoretical yield, actual yield, and percentage yield.

  7. A solid mixture contains \(\ce{Na2SO4}\) and \(\ce{NaCl}\) . What compound would you use to precipitate the \(\ce{Cl-}\) ion?

    Hint: silver ion, \(\ce{Ag+}\)

    Skill -
    Apply what you know of substances. \(\ce{AgCl}\) is one of the few insoluble chlorides.

  8. A solid mixture contains \(\ce{Na2SO4}\) and \(\ce{NaCl}\) . What compound would you use to precipitate the \(\ce{SO4^2-}\) ion?

    Hint: \(\ce{BaCl2}\)

    \(\ce{BaCl2}\) is a soluble salt, but \(\ce{BaSO4}\) is not.

    Skill -
    Apply what you know of substances.

  9. A 2.21 g mixture containing \(\ce{Na2SO4}\) and \(\ce{NaCl}\) dissolved in water is treated with \(\ce{BaCl2}\). The resulting dry \(\ce{BaSO4}\) weighs 2.57 g. How many mole(s) of \(\ce{BaSO4}\) were collected? (formula wt. \(\ce{BaSO4}\), 233.39)

    Hint: 0.0110 mol

    \(\mathrm{\dfrac{2.57\: g}{233.39\: g/mol} =\: ?\: mol}\)

    How many mole(s) of \(\ce{Na2SO4}\) were present in the mixture? (Also 0.0110 mol)

    Skill -
    Performing unit conversion tasks.

  10. A 2.21 g mixture containing \(\ce{Na2SO4}\) and \(\ce{NaCl}\) dissolved in water is treated with \(\ce{BaCl2}\), and analysis shows 0.01101 mol \(\ce{Na2SO4}\) was present. What is the weight of \(\ce{Na2SO4}\)? (formula wt. \(\ce{Na2SO4}\), 142.04)

    Hint: 1.56 g

    Skill -
    Performing unit conversion tasks.

  11. A 2.21 g mixture containing 1.56 g \(\ce{Na2SO4}\) and some unknown impurities is treated with \(\ce{BaCl2}\). How many grams of \(\ce{BaSO4}\) should form? (formula wt. \(\ce{Na2SO4}\), 142.04, \(\ce{BaSO4}\), 233.39)

    Hint: 2.57 g

    Conversion calculation:

    \(\mathrm{1.56\: g\: NaSO_4\times\dfrac{233.39\: g/mol\: BaSO_4}{142.04\: g/mol\: Na_2SO_4}=\: ?\: g\: BaSO_4}\)

    Skill -

    You have acquired all the skills to perform chemical analysis to find out the percentage of certain components in a mixture. For example, consider this problem.

    A 2.21 g mixture containing \(\ce{Na2SO4}\) and \(\ce{NaCl}\) dissolved in water is treated with \(\ce{BaCl2}\), and 2.57 g of \(\ce{BaSO4}\) was obtained. What is the percentage of \(\ce{NaCl}\) in the sample?

  12. Which one of the following samples has the highest percentage of chloride?
    1. pure \(\ce{CaCl2}\)
    2. pure \(\ce{AlCl3}\)
    3. equal weight mixture of \(\ce{CaCl2}\) and \(\ce{AlCl3}\)
    4. pure \(\ce{NaCl}\)
    5. pure \(\ce{BaCl2}\)

    Hint: \(\ce{AlCl3}\)

    Skill -
    Interpret chemical formula

  13. The reaction

    \(\ce{C3H8 + 5 O2 \rightarrow 3 CO2 + 4 H2O}\)

    is best classified as a

    1. formation reaction
    2. decomposition reaction
    3. combustion reaction
    4. exchange reaction
    5. displacement reaction

    Hint: c.

    Skill -
    Describe the major features of chemical reactions.

  14. What chemical (give the formula) would you use in order to determine the percentage of \(\ce{CaCl2}\) in a solid mixture of \(\ce{CaCl2}\) and \(\ce{Ca(NO3)2}\). Assume you can first dissolve the solid in water.

    Hint: silver nitrate

    Skill -
    Apply the rule: salts of nitrates are usually soluble.

  15. A 1.140 g mixture of \(\ce{NaCl}\) and \(\ce{CaCl2}\) dissolved in water is mixed with sufficient solution of \(\ce{AgNO3}\) to give 2.868 g of dry \(\ce{AgCl}\). Calculate the WEIGHT percentage of \(\ce{NaCl}\). (\(\ce{Na}\), 23.0; \(\ce{Cl}\), 35.5; \(\ce{Ca}\), 40; \(\ce{Ag}\), 107.9)

    Hint: 51.3% \(\ce{NaCl}\)

    Skill -
    Technique suggestion:

    \(\mathrm{2.868\: g\: AgCl = 0.0200\: mol\: (Cl^-\: or\: AgCl)}\)

    Assume the mixture contains x g \(\ce{NaCl}\), then (1.140-x) g is \(\ce{CaCl2}\). Thus, we have:

    \(\mathrm{\#\: mol\: NaCl + 2\, \#\: mol\: CaCl_2 = \#\: mol\: Cl^-}\) lead to

    \(\mathrm{\dfrac{x\: g\: NaCl}{58.5\: g /mol}+\dfrac{2 (1.140 - x)\: CaCl_2}{111.1\: g/mol}= 0.0200\: mol}\)

    Solve the equation for x (x = 0.585 g) and calculate the percentage.

  16. A 1.140 g mixture of \(\ce{NaCl}\) and \(\ce{CaCl2}\) dissolved in water is mixed with sufficient solution of \(\ce{AgNO3}\) to give 2.868 g of dry \(\ce{AgCl}\). Calculate the MOLE percentage of \(\ce{NaCl}\). (\(\ce{Na}\), 23.0; \(\ce{Cl}\), 35.5; \(\ce{Ca}\), 40; \(\ce{Ag}\), 107.9)

    Hint: 66% by mole

    Skill -
    Determine the weight and mole percentages of mixtures.