Compounds

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AP Chemistry › Compounds

Questions 1 - 10

What is the oxidation state of manganese () in the polyatomic permanganate anion ()?

Explanation

When assigning oxidation states to elements of a given compound, non-transition metal elements are assigned specific oxidation states corresponding to their group number and valence relative to a complete octet.

Group 1 elements have 1 valence electron and an oxidation state of +1.

Group 2 elements have 2 valence electrons and an oxidation state of +2.

Group 8 elements (the noble gases) have a complete octet, thus their assigned oxidation state is 0.

Group 7 elements (halogens) have 7 valence electrons and an oxidation state of -1.

Group 6 elements such as oxygen have 6 valence electrons and an oxidation state of -2.

Permanganate has 4 oxygen atoms and an overall charge of -1. The oxidation state of the atom may be found by first calculating the combined oxidation state of the oxygen atoms:

and finding the difference between their combined oxidation state and the overall charge of the ion (-1):

What is the oxidation state of manganese () in the polyatomic permanganate anion ()?

Explanation

Group 1 elements have 1 valence electron and an oxidation state of +1.

Group 2 elements have 2 valence electrons and an oxidation state of +2.

Group 8 elements (the noble gases) have a complete octet, thus their assigned oxidation state is 0.

Group 7 elements (halogens) have 7 valence electrons and an oxidation state of -1.

Group 6 elements such as oxygen have 6 valence electrons and an oxidation state of -2.

Permanganate has 4 oxygen atoms and an overall charge of -1. The oxidation state of the atom may be found by first calculating the combined oxidation state of the oxygen atoms:

and finding the difference between their combined oxidation state and the overall charge of the ion (-1):

Calculate the percent by mass of each element in $PbSO_{4}}$ .

Explanation

The total molar mass of lead (II) sulfate is . Lead contributes , sulfur contributes , and oxygen contributes .

The percent by mass of each element in the compound is found by dividing the mass contribution of that element by the total molar mass of the compound.

$Pb: \frac{207.2gPb}{303.265gtotal}=68.33%$

$S: \frac{32.065gS}{303.265gtotal}=10.57%$

$O: \frac{64gO}{303.265gtotal}=21.1%$

Calculate the percent by mass of each element in $PbSO_{4}}$ .

Explanation

The total molar mass of lead (II) sulfate is . Lead contributes , sulfur contributes , and oxygen contributes .

The percent by mass of each element in the compound is found by dividing the mass contribution of that element by the total molar mass of the compound.

$Pb: \frac{207.2gPb}{303.265gtotal}=68.33%$

$S: \frac{32.065gS}{303.265gtotal}=10.57%$

$O: \frac{64gO}{303.265gtotal}=21.1%$

Which of the following is the alcohol functional group?

-NH2

-OH

-COOH

-COH

-OR

Explanation

The alcohol functional group is -OH

A chemist is trying to find the identity of a compound. She knows that the compound has a molecular weight of $180: \frac{g}{mol}$ , and also that it consists of carbon, hydrogen, and oxygen by mass. What are the empirical and molecular formulas, respectively, for this compound?

$CH_{2}O : \text{and}: C_{6}H_{12}O_{6}$

$CH_{2}O : \text{and}: C_{3}H_{6}O_{3}$

$CHO :\text{and}: C_{6}H_{12}O_{6}$

$C_{6}H_{12}O_{6} :\text{and}: CH_{2}O$

It is impossible to tell without more information

Explanation

The simplest way to begin these types of problems is to assume that we are starting out with 100 grams of the sample compound. This allows us to simplify things, since we know the mass percentages of the elements that make up out unknown compound.

$\text{40% carbon in a 100 gram sample}=\text{40 grams carbon}$

$\text{6.67% hydrogen in a 100 gram sample}=\text{6.67 grams hydrogen}$

$\text{53.3% oxygen in a 100 gram sample}=\text{53.3 grams oxygen}$

Next, we must convert amount of each element from grams into moles.

$40\g\ C\left( \frac{1: mol: C}{12: g: C} \right )=3.33:mol:C$

$6.67:g:H\left (\frac{1: mol: H}{1: g: H}\right )=6.67:mol:H$

$53.3:g:O\left ( \frac{1: mol: O}{16: g: O} \right )=3.33:mol:O$

This gives us a formula of $C_{3.33}H_{6.67}O_{3.33}$

Now, we need to simplify this formula by dividing each of the terms by the greatest common denominator, 3.33, which gives us:

$CH_{2}O$

This is our empirical formula. To obtain the molecular formula, we must first find the mass of our empirical formula, which is $12(1) + 1(2) + 16\left ( 1\right ) = 30: g$ .

Next, we need to divide the molecular weight of the compound by the empirical weight to obtain an integer of $\frac{180}{30} = 6$ . Finally, we multiply the empirical formula by this integer in order to obtain the molecular formula, which is:

$6(CH_{2}O) = C_{6}H_{12}O_{6}$

What is the percent by mass of bismuth in the compound $Bi_{2}Te_{3}}$ ?

Explanation

The mass percentage of bismuth in the compound will be equal to the mass of bismuth in one mole of compound divided by the total molar mass of the compound.

Bismuth has a molar mass of $208.98\frac{g}{mol}$ . One mole of the compound would result in two moles of bismuth, a total of 417.96g.

Tellurium has a molar mass of $127.6\frac{g}{mol}$ . One mole of the compound would result in three moles of tellurium, a total of 382.8g.

Add the mass of bismuth and the mass of tellurium per mole to find the total molar mass.

Divide the mass of bismuth by the total molecular mass to find the percent by mass of bismuth in the compound.

$\frac{417.96g}{800.76g}\times100%=52.2%$

Calculate the percent by mass of each element in $Al_{2}(CrO_{4})_{3}$ .

Explanation

The total mass of one mole of aluminum (II) chromate is calculated by:

$2(26.98\frac{g}{mol} Al)+3(51.99\frac{g}{mol} Cr)+12(16\frac{g}{mol}O)=401.95\frac{g}{mol}Al_2(CrO_4)_3$

Aluminum, chromate, and oxygen contribute and respectively. Therefore, we can divide each contribution by the total molecular mass to determine percentages by mass.

$Al: \frac{53.96gAl}{401.95gtotal}=13.42%$

$Cr: \frac{155.97gCr}{401.95gtotal}=38.8%$

$O: \frac{192gO}{401.95gtotal}=47.78%$

What is the percent by mass of bismuth in the compound $Bi_{2}Te_{3}}$ ?

Explanation

The mass percentage of bismuth in the compound will be equal to the mass of bismuth in one mole of compound divided by the total molar mass of the compound.

Bismuth has a molar mass of $208.98\frac{g}{mol}$ . One mole of the compound would result in two moles of bismuth, a total of 417.96g.

Tellurium has a molar mass of $127.6\frac{g}{mol}$ . One mole of the compound would result in three moles of tellurium, a total of 382.8g.

Add the mass of bismuth and the mass of tellurium per mole to find the total molar mass.

Divide the mass of bismuth by the total molecular mass to find the percent by mass of bismuth in the compound.

$\frac{417.96g}{800.76g}\times100%=52.2%$

$CH_{2}O : \text{and}: C_{6}H_{12}O_{6}$

$CH_{2}O : \text{and}: C_{3}H_{6}O_{3}$

$CHO :\text{and}: C_{6}H_{12}O_{6}$

$C_{6}H_{12}O_{6} :\text{and}: CH_{2}O$

It is impossible to tell without more information

Explanation

$\text{40% carbon in a 100 gram sample}=\text{40 grams carbon}$

$\text{6.67% hydrogen in a 100 gram sample}=\text{6.67 grams hydrogen}$

$\text{53.3% oxygen in a 100 gram sample}=\text{53.3 grams oxygen}$

Next, we must convert amount of each element from grams into moles.

$40\g\ C\left( \frac{1: mol: C}{12: g: C} \right )=3.33:mol:C$

$6.67:g:H\left (\frac{1: mol: H}{1: g: H}\right )=6.67:mol:H$

$53.3:g:O\left ( \frac{1: mol: O}{16: g: O} \right )=3.33:mol:O$

This gives us a formula of $C_{3.33}H_{6.67}O_{3.33}$

Now, we need to simplify this formula by dividing each of the terms by the greatest common denominator, 3.33, which gives us:

$CH_{2}O$

This is our empirical formula. To obtain the molecular formula, we must first find the mass of our empirical formula, which is $12(1) + 1(2) + 16\left ( 1\right ) = 30: g$ .

$6(CH_{2}O) = C_{6}H_{12}O_{6}$

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