Question

A 33.153 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 59.060 mg of carbon dioxide and 24.176 mg of water. In another experiment, 34.687 mg of the compound is reacted with excess oxygen to produce 14.99 mg of sulfur dioxide. Add subscripts to the formula provided to correctly identify the empirical formula of this compound. Do not change the order of the elements.

Answer 1

The empirical formula of the compound is C₆H₁₂SO₂.

Explanation:

From the first combustion analysis we know that 33.153 mg of compound will yield 59.060 mg of carbon dioxide (CO₂) and 24.176 mg of water (H₂O).

molecular mass of CO₂ = 44 g/mole

if in         44 mg of CO₂ there are 12 mg of carbon

then in   59.060 mg of CO₂ there are Y mg of carbon

Y = ( 59.060 × 12) / 44 = 16.107 mg of carbon

molecular mass of H₂O = 18 g/mole

if in         18 mg of H₂O there are 2 mg of hydrogen

then in   24.176 mg of H₂O there are X mg of hydrogen

X = (24.176 × 2) / 18 = 2.686 mg of hydrogen

Knowing the quantity of hydrogen and carbon present in the sample, we devise the following reasoning:

if in        33.153 mg of sample there are 16.107 mg of carbon and 2.686 mg of hydrogen

then in  100 mg of sample there are A mg of carbon and C mg of hydrogen

A = (100 × 16.107) / 33.153 = 48.58 % carbon

B = (100 × 2.686) / 33.153 = 8.10 % hydrogen

From the second combustion analysis we know that 34.687 mg of compound will yield 14.99 mg of sulfur dioxide (SO₂).

molecular mass of SO₂ = 64 g/mole

if in         64 mg of SO₂ there are 32 mg of sulfur

then in    14.99 mg of SO₂ there are Z mg of sulfur

Z = (14.99 × 32) / 64 = 7.496 mg of sulfur

Knowing the quantity of sulfur in the sample, we devise the following reasoning:

if in        34.687 mg of sample there are 7.496 mg of sulfur

then in  100 mg of sample there are C mg of sulfur

C = (100 × 7.496) /  34.687 = 21.61 % sulfur

Now the percent of oxygen is determined with the following formula:

% carbon + % carbon hydrogen + % sulfur + % oxygen = 100

% oxygen = 100 - (% carbon + % hydrogen + % sulfur)

% oxygen = 100 - (48.58 + 8.10 + 21.61)

% oxygen = 21.71 %

Knowing the percentages of each element in the compound, following the next algorithm, we determine the empirical formula:

first we divide each percentage with the atomic weight of each element

% carbon / 12 = 48.58 / 12 = 4.048

% hydrogen / 1 = 8.10 / 1 = 8.10

% sulfur / 32 = 21.61 / 32 = 0.67

% oxygen / 16 = 21.71 / 16 = 1.357

now we divide each number by the lowest one, that is 0.67

for carbon = 4.048 / 0.67 ≈ 6

for hydrogen = 8.10 / 0.67 ≈ 12

for sulfur = 0.67 / 0.67 ≈ 1

for oxygen = 1.357 / 0.67 ≈ 2

The empirical formula of the compound is C₆H₁₂SO₂.

Learn more about:

empirical formula

https://brainly.com/question/821181

#learnwithBrainly