Question

1. A compound with a molecular weight of 222 g/mole contains only carbon, hydrogen, and oxygen. When 4.33 mg of this compound was completely combusted in excess oxygen, the products were 9.44 mg of CO2 and 1.75 mg of H2O.

What is the molecular formula of this compound?

2,

Consider this reaction:    2 C2H2 + 5 O2 ? 4 CO2 + 2 H2O

If 222 grams of C2H2 reacts completely in this reaction:

1. What mass of O2 will be required?

2. What mass of CO2 will be formed?

Answer 1

SOLUTION Q1

Given, the compound only contain carbon, oxygen and hydrogen

Let's assume the molecular formula for the compound will be C_xH_YO_z

The reaction of   C_xH_YO_z with excess O₂

C_xH_YO_z +  O₂ CO₂ + H₂O.

Here C_xH_YO_z is excess reagent because  O₂ is in excess

moles of   C_xH_YO_z = mass / molar mass =  4.33 mg / 222 g/mole = 0.01950 mmol

moles of   CO₂ =  mass / molar mass = 9.44 mg / 44g/mole = 0.2145 mmol

moles of  H₂O =  mass / molar mass = 1.75 mg /18g/mole = 0.09722 mmol

Now, calculating moles ratio

moles of   CO₂ / moles of   C_xH_YO_z =  0.2145 mmol / 0.01950 mmol = 11

moles of  H₂O / moles of   C_xH_YO_z = 0.09722 mmol / 0.01950 mmol = 5

So, from 1 mole of C_xH_YO_z we will get 11 mol of  CO₂ and 5 mol of H₂O

Now, balancing the elements in the chemical reaction

1 C_xH_YO_z +  O₂ 11CO₂ + 5H₂O.

balancing the H : Y = 5*2 = 10

balancing the C : X = 11 * 1 = 11

now, the formula for the compound = C₁₁H₁₀O_Z

molecular mass of   C₁₁H₁₀O_Z = 11*12 + 1* 10 + 16* z = 222

16z = 222 - 142 = 80

z = 80/16 = 5

So, the molecular formula for the compound = C₁₁H₁₀O₅ Answer