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MCAT Physical : Stoichiometry and Analytical Chemistry

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Example Questions

Example Question #31 : Molecular Weight, Molecular Formula, And Moles

Compounds can be distinguished from each other by using their molecular weights. The molecular weight of a compound depends on the individual atomic weights of the elements and the amount of each element present in the compound. Consider hexane for example. Hexane has a molecular formula of C_6H_1_4 . This means that it has 6 carbon atoms and 14 hydrogen atoms. To calculate the molecular weight of hexane, we can simply look up the molecular weight of carbon and hydrogen from the periodic table, multiply each molecular weight by the number of atoms (6 for carbon and 14 for hydrogen), and sum the two numbers. The molecular weight of an element is always given in $\frac{grams}{moles}$ . One mole is the defined as the number of atoms in twelve grams of carbon-12.

How many atoms of oxygen are found in 1g of oxygen gas?

Possible Answers:

1.9* 10^2^2

3.8*10^2^4

3.8*10^2^2

1.9*10^2^4

Correct answer:

1.9* 10^2^2

Explanation:

To solve this question, we need to first convert grams to moles, then moles to atoms. The molecular weight (MW) of oxygen gas, O_2 is

$MW\:of\:O_2 = 2\left(16\frac{g}{mol}\right) = 32\frac{g}{mol}$

The amount of moles in 1g of oxygen gas is

$moles\: of \: O_2 = (1g)\left(\frac{1mol}{32g}\right) = 0.0312 \: mol$

There are 6.022* 10^2^3 atoms in $1\:mol$ (this is called the Avogadro’s number); therefore, the number of atoms in of O_2 is

$atoms\ O_2=(0.0312\: mol)\left(\frac{6.022* 10^2^3 \: atoms}{1\: mol}\right)= 1.9*10^2^2\: atoms$

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Example Question #31 : Stoichiometry And Analytical Chemistry

Consider two carbohydrates A and B. Molecule A is a 6-carbon carbohydrate and has twice as much molecular weight as molecule B. What can you conclude about molecule B?

Possible Answers:

None of these

Molecule B has twice as many oxygen atoms as molecule A

Molecule A has twice as many oxygen atoms as carbon atoms

Oxygen contributes to about $53\%$ of the molecular weight of both molecules A and B

Correct answer:

Oxygen contributes to about $53\%$ of the molecular weight of both molecules A and B

Explanation:

The empirical formula for carbohydrates is C_nH_2_nO_n , where is the number of carbon atoms. The question states that molecule A has 6 carbons; therefore, n=6 and the molecular formula for molecule A is C_6H_1_2O_6 . The molecular weight (MW) of this compound is calculated using the molecular weight of each atom. The MW of each atom is obtained from the periodic table and is multiplied by the number atoms.

$(6) \left(12\frac{g}{mol}\right) + (12)\left(1.008\frac{g}{mol}\right) + (6) \left(16\frac{g}{mol}\right) = 180\frac{g}{mol}$

So, the molecular weight of molecule A is $180\frac{g}{mol}$ . Since it is also a carbohydrate, molecule B will have the same empirical formula as molecule A. Molecule B has half the molecular weight of molecule A; therefore, molecule B must have half the atoms as molecule A. The molecular formula of molecule B is C_3H_6 O_3 . Note that MW of is $90\frac{g}{mol}$ .

The MW contribution of oxygen to each compound is calculated as follows.

%MW of oxygen for molecule A = $\frac{(6)(16\frac{g}{mol})}{180\frac{g}{mol}} = 0.53$

%MW of oxygen for molecule B = $\frac{(3)(16\frac{g}{mol})}{90\frac{g}{mol}} = 0.53$

Therefore, oxygen contributes to 53% of MW in both compounds.

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Example Question #1 : Reaction Calculations And Limiting Reagent

What products will be formed by the reaction between copper (II) fluoride and sodium sulfate?

Possible Answers:

Cu₂Na and F₂SO₄

CuS and NaF

Cu₂SO₄ and NaF

CuSO₄ and NaF

Correct answer:

CuSO₄ and NaF

Explanation:

Recognize that this is a double replacement reaction in which ion pairs will switch. After reaction, copper (II) will now be paired with sulfate, and sodium with flouride according to the (unbalanced) equation below. Remember that sulfate has a charge of -2, and flouride of -1.

$CuF_{2 } + Na_{2}SO_{4}\rightarrow CuSO_{4} + NaF$

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Example Question #31 : Stoichiometry And Analytical Chemistry

Which represents the correct balanced equation for the reaction between silver (I) nitrate and magnesium hydroxide?

Possible Answers:

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + 2AgOH$

$2Ag_{3}N + 3Mg(OH)_{2} \rightarrow Mg_{}3N_{}2 + 6AgOH$

$2Ag_{2}(NO_{3}) + Mg(OH)_{2} \rightarrow 2Ag_{2}(OH) + Mg(NO_{3})_{2}$

$2Ag(NO_2)+Mg(OH)_2\rightarrow Mg(NO_2)_2+2AgOH$

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + AgOH$

Correct answer:

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + 2AgOH$

Explanation:

Silver (I) nitrate is AgNO₃. Recognizing this allows us to eliminate two answer choices, which incorrectly substitute nitrogen (N) for nitrate (NO₃) or balance the molecular charges incorrectly. Of the two remaining choices, only one is balanced correctly.

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Example Question #3 : Reaction Calculations And Limiting Reagent

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

How many molecules of carbon monoxide are produced per liter of hydrogen, when using the natural gas method of production at STP?

Possible Answers:

$3*10^{22}\ \text{molecules}$

$6*10^{21}\ \text{molecules}$

$5*10^{24}\ \text{molecules}$

$9*10^{21}\ \text{molecules}$

Correct answer:

$9*10^{21}\ \text{molecules}$

Explanation:

Since the reaction takes place at STP, we can convert liters of hydrogen to moles using a constant.

$1L\ H_2*\frac{1mol}{22.4L}=0.0446mol\ H_2$

Using this value, we can find the moles of carbon monoxide produced and convert to molecules by using Avogadro's number.

$0.0446mol\ H_2*\frac{1mol\ CO}{3mol\ H_2}*\frac{6.022*10^{23}molec}{1mol}=9*10^{21}\ \text{molecules}\ CO$

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Example Question #31 : Stoichiometry And Analytical Chemistry

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

Suppose equal masses of iron and methane were available, along with excess water. What is the difference in the amount of hydrogen molecules that can be produced?

Possible Answers:

Methane will produce about eight times more hydrogen molecules

Iron will produce about eight times more hydrogen molecules

Both methods will produce equal amounts of hydrogen molecules

Iron will produce about 1.3 times more hydrogen molecules

Correct answer:

Methane will produce about eight times more hydrogen molecules

Explanation:

To determine which method will produce more hydrogen molecules, convert equal masses of each reactant into molecules of hydrogen gas. The conversion requires us to convert the initial compound to moles, multiply by the molar ratio from the reaction to find moles of hydrogen, and multiply by Avogadro's number to convert to molecules.

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

$1 g Fe * \frac{1 mol Fe}{55.8 g Fe} * \frac{4 mol H_{2}}{3 mol Fe} * \frac{6.022 * 10^{23} molec H_{2}}{mol H_{2}} = 0.144 * 10^{23} molec$

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

$1 g CH_{4} * \frac{1 mol CH_{4}}{16.0 g CH_{4}} * \frac{3 mol H_{2}}{1 mol CH_{4}} * \frac{6.022 * 10^{23} molec H_{2}}{mol H_{2}} = 1.14 * 10^{23} molec$

Dividing these solutions, we see that the methane method produces roughly eight times the amount of hydrogen molecules.

$\frac{1.14*10^{23}}{0.144*10^{23}}=7.9\approx 8$

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Example Question #5 : Reaction Calculations And Limiting Reagent

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

Suppose $7.7 * 10^{20} atoms$ of iron react with 50g of steam to create iron oxide and hydrogen. What is the limiting reagent and how much of the excess reactant remains?

Possible Answers:

Iron is the limiting reagent and 19g of steam will remain

Steam is the limiting reagent and $5.1 * 10^{15} atoms$ of iron will remain

Iron is a catalyst, and will not be consumed

Iron is the limiting reagent and 49.97g of steam will remain

Correct answer:

Iron is the limiting reagent and 49.97g of steam will remain

Explanation:

In order to identify an excess reactant, we use stoichiometry to convert atoms of iron to mass of steam.

$7.7*10^{20}atoms\ Fe*\frac{1mol\ Fe}{6.022*10^{23}atoms}*\frac{4mol\ H_2O}{3mol\ Fe}*\frac{18.0g\ H_2O}{1mol\ H_2O}=0.031g\ H_2O$

We have 50g of steam available, but only require 0.031g to fully react the given iron. Iron is thus the limiting reagent, since it will be fully consumed first.

50g-0.031g=49.97g

There will be 49.97g of excess steam.

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Example Question #1 : Reaction Calculations And Limiting Reagent

How many milliliters of 0.5M NaOH are required to react completely with 10g of BaCO₃?

Possible Answers:

400mL

100mL

50mL

200mL

Correct answer:

200mL

Explanation:

First, write a balanced equation for the reaction.

$BaCO_{3} + 2NaOH \rightarrow Ba(OH)_{2}) + Na_{2}CO_{3}$

Next, convert 10g BaCO₃ to milliliters of NaOH.

$10 g BaCO_{3} * (\frac{1 mol BaCO_{3}}{197.3 g BaCO_{3}}) * (\frac{2 mol NaOH}{1 mol BaCO_{3}}) * (\frac{1000 mL NaOH}{0.5 mol NaOH}) = 200 mL$

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Example Question #32 : Stoichiometry And Analytical Chemistry

Which of the following produces a $\small 0.1 M$ solution of potassium chloride?

Possible Answers:

Potassium chloride is not soluble in water

Add $\small 74.6g$ of solute to one liter of water

Put $\small 147.51g$ of solute into a container and bring the volume to one liter by adding water while stirring

Put $\small 0.746g$ of solute into a container and bring the volume to $\small 100mL$ by adding water while stirring

Add $\small 7.46g$ of solute to $\small 100mL$ of water

Correct answer:

Put $\small 0.746g$ of solute into a container and bring the volume to $\small 100mL$ by adding water while stirring

Explanation:

We will be looking for a solution that results in one mole of potassium chloride per ten liters of water.

$0.1M=\frac{0.10mol}{1L}$

We will need to find the molecular weight of potassium chloride.

$KCl\rightarrow MW=39.1+35.5=74.6\frac{g}{mol}$

In order to get the desired concentration, we will need to add one-tenth of this amount to one liter of water.

$7.46g*\frac{1mol}{74.6g}=0.1mol$

Our ratio, then is:

$\frac{7.46g}{1L}$

The only answer to follow this ratio is $\small 0.746g$ of potassium chloride in $\small 100mL$ .

$\frac{7.46g\div10}{1L\div 10}=\frac{0.746g}{0.1L}$

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Example Question #32 : Stoichiometry And Analytical Chemistry

$AlCl_{3}+3AgNO_{3}\rightarrow\: Al(NO_{3})_{3}+3AgCl$

If you begin with $\small 66.67g$ of aluminum chloride and unlimited silver nitrate, how many grams of silver chloride can be produced?

Possible Answers:

$186g\ AgCl$

$116g\ AgCl$

$215g\ AgCl$

$35.8g\ AgCl$

$201g\ AgCl$

Correct answer:

$215g\ AgCl$

Explanation:

This is a stoichiometry question requiring us to convert between grams, moles, reactants, and products.

Use the periodic table to find the molar masses of the two compounds in question.

$MW_{AlCl_3}=133.5\frac{g}{mol}$

$MW_{AgCl}=143.37\frac{g}{mol}$

We can use the reaction formula to find the ratio of aluminum chloride to silver chloride. In this case, the ratio is 1:3.

$AlCl_{3}+3AgNO_{3}\rightarrow\: Al(NO_{3})_{3}+3AgCl$

Now we can set up a calculation to convert grams of aluminum chloride to grams of silver chloride, making sure that all units cancel appropriately.

$66.67g\ AlCl_3*\frac{1mol\ AlCl_3}{133.5g\ AlCl_3}*\frac{3mol\ AgCl}{1mol\ AlCl_3}*\frac{143.37g\ AgCl}{1mol\ AgCl}=214.80g\ AgCl$

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MCAT Physical : Stoichiometry and Analytical Chemistry

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #31 : Molecular Weight, Molecular Formula, And Moles

Example Question #31 : Stoichiometry And Analytical Chemistry

Example Question #1 : Reaction Calculations And Limiting Reagent

Example Question #31 : Stoichiometry And Analytical Chemistry

Example Question #3 : Reaction Calculations And Limiting Reagent

Example Question #31 : Stoichiometry And Analytical Chemistry

Example Question #5 : Reaction Calculations And Limiting Reagent

Example Question #1 : Reaction Calculations And Limiting Reagent

Example Question #32 : Stoichiometry And Analytical Chemistry

Example Question #32 : Stoichiometry And Analytical Chemistry

All MCAT Physical Resources

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