Question

(a) The percent composition of an unknown compound is 40.48% C, 34.66% O, 16.86% N, and 8.01% H. What is its empirical formula? Show your work.

(b) What mass of H₂O(l) will we produce when we do the combustion of 15.0 g of C₄H₁₀(l) with an excess of O₂(g) to produce CO₂(g) and H₂O(l)?

Answer 1

b.

Balanced equation for this reaction is ,

2C4H10 + 13O2 ---------> 8CO2 + 10H2O

2 moles of C4H10 gives 10 moles of H2O.

1 mole of C4H10 will give 5 moles of H2O.

Number of moles = Mass/Molar mass

Molar mass of C4H10 = 58.12 g/mol

Moles of C4H10 = 15/58.12

= 0.26 moles

Moles of H2O formed = 0.26 × 5

= 1.30 moles

Molar mass of H2O = 18 g/mol

Mass of H2O formed = 18 g/mol × 1.30 mol

= 23.4 grams of H2O

Answer 2

The weight. This empirical formula. can be calculated from the percentage Composition of the compound as follows:- 1) The percentage weight of each element is divided by it's atomic gives the ratio of the number of atoms. in a molecule of the compound 2.) To get a whole number natio of atoms of different elements, the number in stebe I are divided by the smalles number. 3) The empirical formula is now desived by wouiting the symbols ods of various elements side by side with the number of atoms of each one as the subscript to the lowest sigh of it's symbol

Elements Percent Atomic Ratio Mass. Simplest Ratio C 40.84% 12.01g/mol Imol 40.84 12.01 3.4 1.2 73 N 16:86 » 3.4 14.01 g/mol 16-86 14.01 34.66 >12 1.2 1.2 16.00 g/mol 34-66 16 => 2.2 8.01 101 = 7.9 H 8-01 2.2 1-2 2 1.01 g/mol 7.9 1.2 = Empirical formula → C3H2O,N
b.

Balanced equation for this reaction is ,

2C4H10 + 13O2 ---------> 8CO2 + 10H2O

2 moles of C4H10 gives 10 moles of H2O.

1 mole of C4H10 will give 5 moles of H2O.

Number of moles = Mass/Molar mass

Molar mass of C4H10 = 58.12 g/mol

Moles of C4H10 = 15/58.12

= 0.26 moles

Moles of H2O formed = 0.26 × 5

= 1.30 moles

Molar mass of H2O = 18 g/mol

Mass of H2O formed = 18 g/mol × 1.30 mol

= 23.4 grams of H2O