MCAT Physical - 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 of carbon, of hydrogen, and 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 .

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.

$12\ .\ .\ .\ 2$

$10\ .\ .\ .\ 3$

$12\ .\ .\ .\ 3$

$10\ .\ .\ .\ 2$

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 .

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

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.

$12\ .\ .\ .\ 2$

$10\ .\ .\ .\ 3$

$12\ .\ .\ .\ 3$

$10\ .\ .\ .\ 2$

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.

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

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

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

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

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

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

$2Mn + 10HCl \rightarrow 2MnCl_{5} + 5H_{2}$

5.6 grams of manganese reacts with 650 mL of 6.0 M hydrochloric acid to form manganese (V) chloride and hydrogen gas. Along with the products, a large amount of heat is evolved.

Assuming standard temperature and pressure, what volume of hydrogen gas is produced by this reaction?

Explanation

Convert 5.6g of manganese (limiting reagent) to volume of hydrogen gas.

$5.6 g Mn * (\frac{1 mol Mn}{54.9 g Mn}) * (\frac{5 mol H_{2}}{2 mol Mn}) * (\frac{22.4 L H_{2}}{1 mol H_{2}}) = 5.7 L$

$2Mn + 10HCl \rightarrow 2MnCl_{5} + 5H_{2}$

5.6 grams of manganese reacts with 650 mL of 6.0 M hydrochloric acid to form manganese (V) chloride and hydrogen gas. Along with the products, a large amount of heat is evolved.

Assuming standard temperature and pressure, what volume of hydrogen gas is produced by this reaction?

Explanation

Convert 5.6g of manganese (limiting reagent) to volume of hydrogen gas.

$5.6 g Mn * (\frac{1 mol Mn}{54.9 g Mn}) * (\frac{5 mol H_{2}}{2 mol Mn}) * (\frac{22.4 L H_{2}}{1 mol H_{2}}) = 5.7 L$

The melting point of iron is $2800^{\circ}F$ . In Celsius the melting point of iron is __________ and in Kelvin it is __________.

$^oF=(\frac{9}{5})^oC+32$

$1538^oC\ .\ .\ .\ 1811K$

$1811^oC\ .\ .\ .\ 1538K$

$2400^oC\ .\ .\ .\ 1433K$

$1433^oC\ .\ .\ .\ 2400K$

Explanation

The question states that the melting point of iron is $2800^{\circ}F$ . We can use the given equation to convert this to Celsius.

$^oF=(\frac{9}{5})^oC+32$

$^oC=(^oF-32)(\frac{5}{9})$

$^oC=(2800^oF-32)(\frac{5}{9})$

This means that the melting point of iron is $1538^{\circ}C$ . At this point you can choose the right answer because there is only one answer option that has $1538^{\circ}C$ ; however, let’s solve for Kelvin. The second equation you need to know is the conversion from Celsius to Kelvin:

The melting point of iron in Kelvin is:

The melting point of iron is $2800^{\circ}F$ . In Celsius the melting point of iron is __________ and in Kelvin it is __________.

$^oF=(\frac{9}{5})^oC+32$

$1538^oC\ .\ .\ .\ 1811K$

$1811^oC\ .\ .\ .\ 1538K$

$2400^oC\ .\ .\ .\ 1433K$

$1433^oC\ .\ .\ .\ 2400K$

Explanation

The question states that the melting point of iron is $2800^{\circ}F$ . We can use the given equation to convert this to Celsius.

$^oF=(\frac{9}{5})^oC+32$

$^oC=(^oF-32)(\frac{5}{9})$

$^oC=(2800^oF-32)(\frac{5}{9})$

The melting point of iron in Kelvin is:

How many sodium ions are present in $\small 100mL$ of a $\small 0.023 mM$ solution of sodium hydroxide?

$\small 1.385*10^{18}$

$\small 1.385*10^{19}$

$\small 2.6*10^{-12}$

$\small 2.3 * 10^{-8}$

$\small 2.6*10^{18}$

Explanation

A full liter of a one molar solution of sodium hydroxide would contain one mole of sodium ions, or $\small 6.022*10^{23}$ ions. Here, you have only one tenth the volume, so multiply the number in one mole by one tenth.

$\small 10^{-1}*(6.022*10^{23 })= 6.022*10^{22}ions$

Now that we have reduced the volume, we need to account for the concentration.

$0.023mM=0.023*10^{-3}M$

$(6.022*10^{22}ions)*(0.023*10^{-3}M)=1.385*10^{18}ions$

How many sodium ions are present in $\small 100mL$ of a $\small 0.023 mM$ solution of sodium hydroxide?

$\small 1.385*10^{18}$

$\small 1.385*10^{19}$

$\small 2.6*10^{-12}$

$\small 2.3 * 10^{-8}$

$\small 2.6*10^{18}$

Explanation

$\small 10^{-1}*(6.022*10^{23 })= 6.022*10^{22}ions$

Now that we have reduced the volume, we need to account for the concentration.

$0.023mM=0.023*10^{-3}M$

$(6.022*10^{22}ions)*(0.023*10^{-3}M)=1.385*10^{18}ions$

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

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

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

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

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

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

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.

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

When calculating the empirical formula, if you used ratios of the number of atoms of each element instead of ratios of moles of each element, would you get a different answer?

No, because there is a constant relationship between moles and atoms of each element

No, because the number of moles and the number of atoms in each element is equal

Yes, because there is a constant relationship between mass and atoms of each element

Yes, because Avogadro’s number is different for each element

Explanation

Remember that the empirical formula relies on the ratio of the moles of elements. To get the number of atoms, you would have to multiply the moles of each element by the Avogadro’s number ( $6.022*10^{23}$ ). You would use this number for every element (Avogadro’s number doesn’t change for each element). This means that the ratio of the number of atoms will be the same as the ratio of the number of moles, and you will get the same empirical formula. There is a constant relationship between the number of moles and the number of atoms in a sample.

Stoichiometry and Analytical Chemistry

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