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MCAT Physical : Stoichiometry and Analytical Chemistry

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Example Question #11 : Stoichiometry And Analytical Chemistry

What is the empirical formula of 4-octene?

Possible Answers:

CH_2

$C_8H_{16}$

$C_8H_{15}$

C_2H_7

C_4H_8

Correct answer:

CH_2

Explanation:

4-octene looks like this:

CH_3CH_2CH_2CH=CHCH_2CH_2CH_3

To get an empirical formula, we find the ratio of each element within the compound and make it as low as possible. We have eight carbons and sixteen hydrogens. The ratio of carbons to hydrogens is 8-to-16, which reduces to 1-to-2. The full formula for 4-octene is $C_8H_{16}$ , and the empirical formula is CH_2 .

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Example Question #11 : Stoichiometry And Analytical Chemistry

Which of the following answer choices is not written as an empirical formula?

Possible Answers:

C_4H_7NO_4

$C_{6}H_{12}O_{6}$

$KClO_{4}$

Na_2O

$C_{2}H_{4}O$

Correct answer:

$C_{6}H_{12}O_{6}$

Explanation:

An empirical formula must be written as the most simplified ratio of the elements that the compound contains. For example, $KClO_{4}$ is empirical because it cannot be simplified any further; the ratio of its atoms is 1:1:4.

The formula for glucose, $\small C_6H_{12}O_6$ , can be simplified by a factor of six. The empirical formula for glucose would be CH_2O .

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Example Question #13 : Stoichiometry And Analytical Chemistry

A researcher performs an elemental analysis on a compound. He finds that the compound is made up of only carbon, hydrogen, and oxygen atoms. He isolates a pure sample of the compound and finds that this sample contains 144g of carbon, 24g of hydrogen, and 32g of oxygen. The researcher wants to perform further analysis on this compound the next day. Before leaving the lab the researcher creates three stock solutions of varying concentrations of this compound: (solution A), (solution B), and (solution C). He stores these solutions overnight at a temperature of 15^oC .

Molecular weight of this compound = $300.486\frac{g}{mol}$

Compared to the empirical formula, the molecular formula contains __________ more atoms of carbon and __________ more atoms of oxygen.

Possible Answers:

$12\ .\ .\ .\ 3$

$10\ .\ .\ .\ 3$

$10\ .\ .\ .\ 2$

$12\ .\ .\ .\ 2$

Correct answer:

$12\ .\ .\ .\ 2$

Explanation:

The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.

$(144g)(\frac{1mol}{12.001g})=12\ mol\ C$

$(24g)(\frac{1mol}{1.008g})=24\ mol\ H$

$(32g)(\frac{1mol}{16g})=2\ mol\ O$

After finding the moles of each element, you have to find the smallest whole number ratio of each element. The smallest whole number ratio can be found by dividing moles of each element by the lowest mole quantity (in this case, $2\ mol$ of oxygen). You are left with carbons, hydrogens, and oxygen. The empirical formula for this compound is C_6H_1_2O .

To find the molecular formula of the compound you need to divide the molecular weight of the actual compound by the molecular weight of the empirical formula. The molecular weight of the empirical formula is:

$(6*12.001\frac{g}{mol})+(12*1.008\frac{g}{mol})+(1*16\frac{g}{mol})=100\frac{g}{mol}$

Dividing the molecular weight of the actual compound ( $300.486\frac{g}{mol}$ ) by the molecular weight of empirical formula gives:

$\frac{300.486\frac{g}{mol}}{100\frac{g}{mol}} = 3$

This means that the empirical formula must be multiplied by three to get the molecular formula; therefore, the molecular formula is C_1_8H_3_6O_3 . Compared to the empirical formula, the molecular formula contains more carbon atoms and more oxygen atoms.

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Example Question #14 : Stoichiometry And Analytical Chemistry

Molecular weight of this compound = $300.486\frac{g}{mol}$

What is the empirical formula of this compound?

Possible Answers:

C_8H_1_6O

C_6H_1_2O

C_5H_1_0O

C_7H_1_4O

Correct answer:

C_6H_1_2O

Explanation:

The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.

The molecular weight of carbon is $12.001\frac{g}{mol}$ , hydrogen is $1.008\frac{g}{mol}$ , and oxygen is $16\frac{g}{mol}$ .

The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.

$(144g)(\frac{1mol}{12.001g})=12\ mol\ C$

$(24g)(\frac{1mol}{1.008g})=24\ mol\ H$

$(32g)(\frac{1mol}{16g})=2\ mol\ O$

$\frac{12\ mol\ C}{2\ mol}=6\ mol\ C$

$\frac{24\ mol\ H}{2\ mol}=12\ mol\ H$

$\frac{2\ mol\ O}{2\ mol}=1\ mol\ O$

You are left with carbons, hydrogens, and oxygen. The empirical formula for this compound is C_6H_1_2O .

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Example Question #15 : Stoichiometry And Analytical Chemistry

Which of the following is both an empirical formula and a molecular formula?

Possible Answers:

$C_{4}H_{6}O_{2}$

$C_{6}H_{12}O_{6}$

$C_{2}H_{4}O_{2}$

$C_{3}H_{6}O_{9}$

$C_{2}H_{4}O$

Correct answer:

$C_{2}H_{4}O$

Explanation:

An empirical formula is the simplest form of a molecular formula that still retains the ratio of the elements. If a formula can be divided by a whole number, it is a molecular formula and not an empirical formula. A molecular formula is the exact identity of a compound, showing the total number of atoms used to create the compound.

The only given answer that is not divisible by a whole number is C_2H_4O , making it an empirical formula. It is also the molecular formula for both acetaldehyde and ethanol, depending on molecular geometry and orientation.

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Example Question #11 : Molecular Weight, Molecular Formula, And Moles

Which of the following is a molecular formula but not an empirical formula?

Possible Answers:

$C_{3}H_{6}O_{2}$

$C_{3}H_{6}O_{}$

$CH_{2}O$

$C_{6}H_{12}O_{6}$

$C_{4}H_{10}O_{3}$

Correct answer:

$C_{6}H_{12}O_{6}$

Explanation:

The only given answer that is fully divisible by a whole number is $C_6H_{12}O_6$ . This means it is not an empirical formula. Any given formula, however, will be a molecular formula, provided that the atoms are capable of forming the appropriate bonds. This is the molecular formula for glucose, which reduces to the empirical formula of CH_2O . CH_2O is the empirical formula for a generalized carbohydrate molecule.

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Example Question #11 : Stoichiometry And Analytical Chemistry

A given compound is composed of $92.3\%$ carbon and $7.7\%$ hydrogen and has a molar mass of $\small 78.0\frac{g}{mol}$ . What are the empirical and molecular formulas of this compound, respectively?

Possible Answers:

$C_6H_6\ \text{and}\ CH$

$CH\ \text{and}\ C_6H_6$

$CH\ \text{and}\ C_4H_8$

$C_6H_6\ \text{and}\ C_2H_2$

Correct answer:

$CH\ \text{and}\ C_6H_6$

Explanation:

To find the empirical formula, use the mass percentage of each element to find mole ratios based on a hypothetical sample of 100g .

$92.3g\ C*\frac{1mol}{12.0g\ C}=7.7mol\ C$

$7.7g\ H*\frac{1mol}{1.00g\ H}=7.7mol\ H$

We see that there is a 1:1 mole ratio for carbon to hydrogen, making the empirical formula .

The next step will be to find molecular formula by dividing the molar mass of the compound by the molar mass of the empirical formula.

$MM_{\text{empirical}}=12.0g+1.00g=13.0\frac{g}{mol}$

$MM_{\text{compound}}=78.0\frac{g}{mol}$

$\frac{MM_{\text{compound}}}{MM_{\text{empirical}}}=\frac{78.0\frac{g}{mol}}{13.0\frac{g}{mol}}=6$

Multiply the subscripts of the empirical formula for each element by six to get the molecular formula: .

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Example Question #18 : Stoichiometry And Analytical Chemistry

A compound is found to have a molar mass of $102 \frac{g}{mol}$ and is composed of $58.8\%$ carbon, $9.8\%$ hydrogen, and $31.4\%$ oxygen. What are the empirical and molecular formulas for the compound, respectively?

Possible Answers:

$CH_{2}O, C_{2}H_{4}O_{2}$

$C_{5}H_{10}O_{2}, C_{5}H_{10}O_{2}$

$C_{5}H_{10}O, C_{2}H_{4}O_{2}$

$C_{5}H_{10}O_{2}, C_{10}H_{20}O_{4}$

Correct answer:

$C_{5}H_{10}O_{2}, C_{5}H_{10}O_{2}$

Explanation:

To find the empirical formula, we must first find the mole ratios by using molar masses for each element:
$58.8 g C * \frac{1 mol C}{12.0 g} = 4.9 mol$

$9.8 g H * \frac{1 mol H}{1.0 g} = 9.8 mol$

$31.4 g O * \frac{1 mol O}{16.0 g} = 1.9 mol$

Rounding these numbers, we get five moles of carbon, ten moles of hydrogen, and two moles of oxygen, making the formula $C_5H_{10}O_2$ . The formula cannot be simplified by any common denominator, and thus represents the final empirical formula.

We know the molecular mass, given in the question. To find the molecular formula, we need to find the ratio of the mass of the empirical formula to the molecular mass.

$C_5H_{10}O_2:\ 5(12)+10(1)+2(16)=102\frac{g}{mol}$

The mass of the empirical formula is equal to the molecular mass, meaning that the empirical and molecular formulas are the same.

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Example Question #11 : Stoichiometry And Analytical Chemistry

Consider the following molecular formulas:

$\text{Ribose: }C_5H_{10}O_5$

$\text{Ethyl butyrate: }C_6H_{12}O_2$

$\text{Chlorophyll: }C_{55}H_{72}MgN_4O_5$

$\text{DEET*: }C_{12}H_{17}ON$

*The IUPAC name for DEET is N,N-diethyl-meta-toluamide

Which of the following is NOT true?

Possible Answers:

The combined mass percentage of hydrogen and carbon is ribose is greater than the mass percentage of oxygen

Oxygen represents a larger percentage by mass in ethyl butyrate than in chlorophyll

The molecular formula for DEET is the same as its empirical formula

The empirical formula for ribose is CH_2O

Correct answer:

The combined mass percentage of hydrogen and carbon is ribose is greater than the mass percentage of oxygen

Explanation:

The molecular formula is the same as the empirical formula if it cannot be reduced by any whole number. Any formula containing a single atom of any given element must be an empirical formula as well. The formula for DEET is $\small C_{12}H_{17}ON$ . Since this contains a single atom each of oxygen and nitrogen, it cannot be further reduced and must be an empirical formula as well.

The molecular formula for ribose is $\small C_5H_{10}O_5$ , which can be reduced by a factor of five. The empirical formula for ribose (and most other carbohydrates) is CH_2O .

The other two options require us to calculate pass percentages based on the given molecular formulas.

Although we can look at the formulas for chlorophyll and ethyl butyrate to deduce that oxygen makes up a larger percentage of the latter, we can double check mathematically. In order to find which compound contains oxygen in a larger percentage, divide the molar mass of the oxygen in the compound by the molar mass of the entire compound.

For ethyl butyrate:

$C_6H_{12}O_2$

$\frac{2(16\frac{g}{mol})}{6(12\frac{g}{mol})+12(1\frac{g}{mol})+2(16\frac{g}{mol})}=\frac{32.0\frac{g}{mol}}{116\frac{g}{mol}}=0.276$

For chlorophyll:

$C_{55}H_{72}MgN_4O_5$

$\frac{5(16\frac{g}{mol})}{55(12\frac{g}{mol})+72(1\frac{g}{mol})+1(24\frac{g}{mol})+4(14\frac{g}{mol})+5(16\frac{g}{mol})}=\frac{80\frac{g}{mol}}{892\frac{g}{mol}}=0.0896$

We see that it is in fact true that there is a larger percentage of oxygen in ethyl buyrate.

Next, find out if hydrogen and carbon make up the largest mass percentage of ribose by using the same method:

$C_5H_{10}O_5$

$\frac{5(12\frac{g}{mol})+10(1\frac{g}{mol})}{5(12\frac{g}{mol})+10(1\frac{g}{mol})+5(16\frac{g}{mol})}=\frac{70\frac{g}{mol}}{150\frac{g}{mol}}=0.467$

This amounts to 46.7% of the molecular mass, meaning that oxygen must account for the remaining 53.3%. Oxygen thus makes up a greater mass percentage of ribose than hydrogen and carbon combined.

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Example Question #641 : Mcat Physical Sciences

What is the empirical formula for a compound that is composed of 53.3% oxygen, 6.7% hydrogen, and 40% carbon?

Possible Answers:

$CH_{2}O$

$C_{4}HO_{5}$

$C_{4}H_{8}O_{4}$

$CH_{5}O_{3}$

Correct answer:

$CH_{2}O$

Explanation:

Given the percentages by mass of the compound, we can convert the percentages to grams by considering a 100-gram sample of the compound. The next step involves dividing each mass by the element's molar mass.

$\frac{40g}{12\frac{g}{mol}}=3.33mol\ C$

$\frac{6.7g}{1\frac{g}{mol}}=6.7mol\ H$

$\frac{53.33g}{12\frac{g}{mol}}=3.33mol\ O$

By dividing each value by the smallest molar quantity, we can determine the ratio of elements in the compound.

$\frac{3.33}{3.33}mol\ C:\frac{6.7}{3.33}mol\ H:\frac{3.33}{3.33}mol\ O$

$1mol\ C:2mol\ H:1mol\ O$

As a result, the empirical formula for the compound is $CH_{2}O$ .

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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 #11 : Stoichiometry And Analytical Chemistry

Example Question #11 : Stoichiometry And Analytical Chemistry

Example Question #13 : Stoichiometry And Analytical Chemistry

Example Question #14 : Stoichiometry And Analytical Chemistry

Example Question #15 : Stoichiometry And Analytical Chemistry

Example Question #11 : Molecular Weight, Molecular Formula, And Moles

Example Question #11 : Stoichiometry And Analytical Chemistry

Example Question #18 : Stoichiometry And Analytical Chemistry

Example Question #11 : Stoichiometry And Analytical Chemistry

Example Question #641 : Mcat Physical Sciences

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