Stoichiometry - SAT Subject Test in Chemistry

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Question

How many nitrate ions are in one mole of $Ca(NO_{3})_{2}$ ?

Answer

Note that there are two nitrate ( $NO_{3}$ ) ions for every formula unit of calcium nitrate. This means that in one mole of calcium nitrate, there are two moles of nitrate ions. To go from moles to ions, multiply 2 by Avogadro's number.

$2\cdot6.0\cdot10^{23} = 1.2\cdot10^{24}$

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Question

How many hydrogen atoms are in one mole of $(NH_{4})_{2}S$ ?

Answer

In one molecule of $(NH_{4})_{2}S$ , there are 2 ammonium ions, each of which has 4 hydrogen atoms for a total of 8 hydrogen atoms. Thus, to find the number of hydrogen atoms in one mole of $(NH_{4})_{2}S$ , one must multiply Avogadro's number by 8.

$8(6.0\cdot10^{23})=4.8\cdot10^{24}$

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Question

Calculate the number of aluminum ions in $4.5\times10^{-2}$ moles of $\text{Al}_2(\text{Cr}_2\text{O}_7)_3$ .

Answer

In order calculate the number of aluminum ions, we must first find the number of aluminum ions in the entire compound. In this case, there are two molecules of $\text{Al}^{3+}$ for every molecule of $\text{Al}_2(\text{Cr}_2\text{O}_7)_3$ . After understanding this, we can use Avogadro's constant to determine the number of atoms (more specifically ions) of aluminum. See equation below for specific calculations.

$4.5\times10^{-2},\text{mol Al}_2(\text{Cr}_2\text{O}_7)_3\times\frac{2;\text{mol Al}^{3+}}{1,\text{mol Al}_2(\text{Cr}_2\text{O}_7)_3}$ $\times\frac{6.02\times10^{23}; \text{Al}^{3+},\text{ions}}{1;\text{mol Al}^{3+}}=5.4\times10^{22}; \text{Al}^{3+},\text{ions}$

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Question

How many atoms of chloride are in 0.2550 g of aluminum chloride, $\text{AlCl}_3$ ?

Answer

The formula for aluminum chloride is $\text{AlCl}_3$ . The molar mass of $\text{Al}$ is 26.98 g and the molar mass of $\text{Cl}$ is 35.45 g. There are three $\text{Cl}$ atoms in $\text{AlCl}_3$ . To calculate the molecular mass of $\text{AlCl}_3$ , we need to find the sum of the mass of one aluminum atom and three chlorine atoms:

$1(26.98\text{ amu Al})+3(35.45\text{ amu Cl})=133.33\text{ amu}$

The total molecular weight is 130.33 g. Starting with the grams of $\text{AlCl}_3$ , we convert to the amount of moles $\text{AlCl}_3$ using the molecular weight value we just calculated. Then we calculate the number of moles of chloride ion in $\text{AlCl}_3$ . using the 3:1 ratio. Finally, we calculate the number of atoms by using Avogadro's number. See calculations below:

$0.2550\text{ g AlCl}_3 \times \frac{1\text{ mol AlCl}_3}{133.33\text{ g AlCl}_3} \times\frac{3\text{ mol Cl}}{1\text{ mol AlCl}_3}$ $\times\frac{6.02\times10^{23}\text{ atoms Cl}}{1\text{ mol Cl}},= 3.454\times10^{21}\text{ atoms Cl}$

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Question

When the equation for the reaction shown below is balanced with the lowest whole-number coefficients, what is the sum of the coefficients of the reactants?

___ $C_{4}H_{10} +$ ___ $O_{2}$ $\rightarrow$ ___ $H_{2}O +$ ___ $CO_{2}$

Answer

The coefficients of the equation, when balanced, should be 2, 13, 10, and 8. This problem wants to know the sum of the reactants, so only the coefficients on the left side of the reactants should be added. The sum of 2 and 13 is 15, so 15 is the correct answer.

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Question

Which pair of formulas represents the empirical formula and molecular formula for a certain compound?

Answer

To find an empirical formula, take a molecular formula and divide the subscript of each element by the greatest common factor of all the subscripts. In this case, the only pair that works is $CH_2O,C_6H_{12}O_6$ , which can be verified by dividing the coefficients of the molecular formula by 6. Note that is not correct because neither of the formulas is an empirical formula; they are both possible molecular formulas for a compound with the empirical formula .

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Question

Vanillin is composed of carbon, oxygen, and hydrogen.

What is the empirical formula of vanillin?

Refer to the following table for the atomic masses of the elements shown:

$\begin{tabular}{l | c} H & 1.008 amu \ C & 12.01 amu \ O & 16.00 amu \ \end{tabular}$

Answer

In order to calculate the empirical formula of vanillin from the given percent composition data, we can pretend we have a 100 g sample vanillin and from there estimate how many grams of each element make up the sample. For example, Vanillin is carbon; therefore, the estimated sample would contain 66.3 g of carbon. After doing this for each of the elements vanillin contains, we can use the atomic mass of carbon to find how many moles of carbon make up the sample. We then do the same for each element that composes vanillin. See calculations below:

$63.15,\text{g C}\times, \frac{1,\text{mol C}}{12.01,\text{g C}},=5.258;\text{mol C}$

$31.55,\text{g O}\times, \frac{1,\text{mol O}}{16.00,\text{g O}},=1.972;\text{mol O}$

$5.30,\text{g H}\times, \frac{1,\text{mol H}}{1.008,\text{g H}},=5.258;\text{mol H}$

After finding the number of moles of each element in the hypothetical 100 g sample, we look at the ratio between the moles of each element by dividing all the samples by smallest number of moles. After finding the ratios of the samples, we are able to use the empirical formula of the vanillin by rounding up to the nearest whole number for each sample. If the ratios are not close enough to round up (like in this example), we multiply the ratios all by the same number until each number is equal to a whole number nicely. See calculations below:

$\frac{5.258 \text{ mol C}}{1.972\text{ mol O}} = 2.666\text{ mol C}=2\frac{2}{3}\text{ mol C}$

$\frac{5.258 \text{ mol H}}{1.972\text{ mol O}} = 2.666\text{ mol H}=2\frac{2}{3}\text{ mol H}$

$\frac{1.972 \text{ mol O}}{1.972\text{ mol O}} = 1.000\text{ mol O}$

We need whole numbers, so multiply each of the resulting numbers by 3 to get rid of the fractions.

$2\frac{2}{3}\text{ mol C}\times3=8\text{ mol C}$

$2\frac{2}{3}\text{ mol H}\times3=8\text{ mol H}$

$1.000\text{ mol O}\times3=3.000\text{ mol O}$

This gives us $\text{C}_8\text{H}_8\text{O}_3$ as the final answer.

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Question

$2\text{H}_2 +\text{O}_2\rightarrow 2\text{H}_2\text{O}$

Suppose that 8 grams of of hydrogen gas and 16 grams of oxygen gas are ignited. What is the theoretical yield, in grams, of water? Assume hydrogen has an atomic mass of 1 and oxygen has an atomic mass of 16.

Answer

First, convert from grams to moles using the molar masses of the compounds. There are 4 mol of hydrogen gas and 0.5 mol of oxygen gas. Oxygen gas is clearly the limiting reagent (since there is more than twice as much hydrogen gas as oxygen gas), which means that the theoretical yield of water is

$0.5\text{ mol} \times 2 = 1 \text{ mol}$

Water contains two hydrogen atoms and an oxygen atom, so it has a molar mass of

$(2\times1\frac{\text{g}}{\text{mol}}) + (1\times16\frac{\text{g}}{\text{mol}}) = 18\frac{\text{g}}{\text{mol}}$

Hence, 1 mol of water would have a mass of 18 g.

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Question

What is the approximate percentage of carbon by mass in heptane, which has the molecular formula $C_7H_{16}$ ?

Answer

First, calculate the molar mass of heptane, which is

$7(12\text{ amu}) + 16(1\text{ amu}) = 100\text{ amu or } \frac{\text{g}}{\text{mol}}$

We can calculate the percentage of carbon by mass in heptane by creating a fraction of the mass of carbon over the mass of the entire molecule and converting it into a percentage:

$\frac{\text{Mass of C in C}_7\text{H}_{16}}{\text{Mass of C}_7\text{H}_{16}}=\frac{7(12\text{ amu})}{100\text{ amu}}=\frac{84}{100}\times100%=84%\text{ C}$

For every 100 g of heptane, 84 g must come from carbon. This means that the percentage of carbon by mass is

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Question

What is the mass percent of chromium in $\text{Al}_2(\text{Cr}_2\text{O}_7)_3$ ?

Use the values in the following table for the atomic masses of the elements shown.

$\begin{tabular}{l | c} Cr & 52.00 amu \ Al & 26.98 amu \ O & 16.00 amu \ \end{tabular}$

Answer

In order to find the mass percent of a certain atom in a particular molecule, you must know the mass the atom contributes to the molecule and the total mass of the molecule. In this case we calculate the mass percent using the equation

$\frac{\text{Mass of Cr in Al}_2(\text{Cr}_2\text{O}_7)_3}{\text{Atomic Mass of Al}_2(\text{Cr}_2\text{O}_7)_3} \cdot100%$

Calculating the mass of $\text{Cr}$ in the molecule is done by multiplying the atomic mass of a $\text{Cr}$ atom by the number of atoms of that element present in the molecule. There are six $\text{Cr}$ atoms in the presented molecule. We can tell this because $\text{Cr}$ has a subscript of 2 and is contained in a set of parentheses that has a subscript of 3. This tells us that there are three sets of two $\text{Cr}$ atoms in each of the molecules. The atomic mass of a $\text{Cr}$ atom is given as 52.00 amu, so we can multiply:

$6; \text{Cr atoms} \cdot 52.00 \frac{\text{g}}{\text{mol}} = 312.00 \frac{\text{g}}{\text{mol}},\text{Cr}$

Now we need to calculate the total atomic mass of the molecule. We can do this by identifying the atomic mass of each element in the molecule, multiplying it by the number of that type of atom in the molecule, and summing the results together.

$2(26.98 \frac{\text{g}}{\text{mol}};\text{Al}) + 6 (52.00 \frac{\text{g}}{\text{mol}}; \text{Cr}) +21 (16.00 \frac{\text{g}}{\text{mol}}; \text{O}) = 701.96 \frac{\text{g}}{\text{mol}}; \text{Al}_2(\text{Cr}_2\text{O}_7)_3$

We then plug in the values into equation to find the mass percent.

$\frac{311.94 \frac{\text{g}}{\text{mol}}; \text{Cr}}{701.9 \frac{\text{g}}{\text{mol}}; \text{Al}_2(\text{Cr}_2\text{O}_7)_3}\cdot100%=44.4%$

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Question

Octane, $C_{8}H_{18}$ , is an important of component of gasoline, which can be burned as fuel in cars. The following is an unbalanced equation for the combustion of octane.

If 114 g of octane are burned in the complete combustion reaction shown below, how many grams of water will be produced?

___ $C_{8}H_{18} +$ ___ $O_{2} \rightarrow$ ___ $CO_{2}$ ___ $H_{2}O$

Answer

First, the equation should be balanced. This can be done with the coefficients 2, 25, 16, and 18.

Next, calculating the molar mass of octane shows that it has a molar mass of 114 g/mol. This means that 1 mole of octane is used in the reaction, and 9 times as many moles of water should be produced, meaning one of the products will be 9 mol of water. Water has a molar mass of 18 g/mol, so the total mass of water produced is $9\cdot18=162\textup g$ .

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Question

Suppose that 4 moles of undergoes complete decomposition. At STP, how many liters of would be produced as product? (At STP, one mole of a gas takes up 22.4 L of space.)

Answer

Start by creating a skeleton equation for the decomposition reaction:

$NaClO_3 \rightarrow NaCl + O_2$

This can be balanced to give:

$2NaClO_3 \rightarrow 2NaCl + 3O_2$

This means that 4 mol of reactant would produce 6 mol of oxygen gas. The question states that at STP, one mole of a gas takes up 22.4 L of space, so it is easy to find that 6 moles of oxygen gas would occupy a volume of 134.4 L:

$6\text{ mol O}\times\frac{22.4 L}{1 \text{ mol}}=134.4\text{ L}$

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