2. Carbon, hydrogen, and oxygen are not the only elements that can be characteri

Question

2. Carbon, hydrogen, and oxygen are not the only elements that can be characterized by combustion analysis. If a compound also contains sulfur or nitrogen, then it will form CO 2 (from the carbon), H2 O (from the hydrogen), N2 (from any nitrogen), and SO 2 (from any sulfur). The amount of oxygen in the original sample is determined from subtracting the masses of the other elements from the total, as in the combustion analysis described earlier. If a 0.500 g sample of a compound yields 0.814 g CO 2 , 0.204 g H 2 O, 0.0288 g N2 , and 0.132 g SO 2 when burned, then what is its empirical formula? Give your answer in the form C#H#N#O#S#, where the number is the subscript.

3. A mixture of potassium bromide and potassium hydroxide has a total mass of 5.50 g. If the mixture contains 2.45 g K, what is the mass in grams of potassium hydroxide in the mixture? Give the numerical part of your answer only; don’t include the units. 5. Some compounds will decompose (that is, break apart into smaller molecules) when heated, instead of burning. For example, when mercury(II) oxide is exposed to heat it breaks down into mercury and oxygen gas: 2HgO(s) 2Hg(l) + O_2 2 (g). This reaction was one of the critical experiments that lead to the discovery of oxygen gas. As with the other analyses, the important step to using this kind of data to determine empirical formula is to convert the grams of the product compounds into moles of the individual elements. If a 1.75 g sample of an unknown compound decomposes to produce 1.06 g Cr2 O3 , 0.495 g H2 O, and 0.195 g N2 , then what is its empirical formula?

Explanation / Answer

2.

First calculate % composition

% of C

Molar mass of CO2 = 44g/mol

Molar mass of carbon = 12g/mol

that mean 44 gm of CO2 contain 12 gm of C then 0.814 gm CO2 contain 0.814 X 12/44 = 0.222 gm carbon.

Sample is 0.500gm = 100% then 0.222 gm C = 0.222X100/0.500 = 44.4%

% of C = 44.4%

% of H

molar mass of H2O = 18gm/mole and molar mass of H2 = 2gm/mole that mean 18 gm H2O contain 2 gm H2 then 0.204 gm  H2O contain 0.204X 2/18 = 0.0226 gm hydrogen

0.500 gm of sample = 100% then 0.0226 gm = 0.0226 X 100/0.500 = 4.53 %

% of hydrogen = 4.53%

% of Sulfur

Molar mass of SO2 = 64g/mol

Molar mass of S = 32g/mol

that mean 64 gm SO2 contain 32 gm of S then 0.132 gm SO2 contain 32 X 0.132 /64 = 0.066 gm S

Sample is 0.500 gm = 100% then 0.066 gm S = 0.066 X 100/0.500 = 1.2 %S

% of S = 13.2%

% of N

N2 contain both N gm of N = 0.0288 gm

0.500 gm of sample = 100% then 0.0288 gm = 0.0288 X 100 / 0.500 = 5.76 %

% of O = 100 - 44.4 - 4.53 - 13.2 - 5.76 = 32.11%

Calculate atomic ratio by dividing % composition by atomic mass

Atomic ratio of C = 44.4/12 = 3.7

Atomic ratio of H = 4.53/1 = 4.5

Atomic ratio of S = 13.2/32 = 0.4

Atomic retio of N = 5.76 / 14 = 0.4

Atomic ratio of O = 32.11/16 = 2.0

Calculate simplest atomic ratio by dividing atomic ratio by smallest atomic ratio

Simplest atomic ratio of C = 3.7/0.4 = 9.25

Simplest atomic ratio of H = 4.5/0.4 = 11.25

Simplest atomic ratio of S = 0.4/0.4 = 1

Simplest atomic ratio of N = 0.4/0.4 = 1

Simplest atomic ratio of O = 2/0.4 = 5

Convert fractional number to whole number multiplying by

C = 9.25 X 4  = 37

H = 11.25 X 4 = 45

S = 1 X 4 = 4

N = 1 x 4 = 4

O = 5 X 4 = 20

Empirical formula of compound = C37H45N4O20S4

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