Question

(a) The percent composition of an unknown compound is 40.48% C, 34.66% O, 16.86% N, and...

(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)?

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Answer #1

The weight. This empirical formula. can be calculated from the percentage Composition of the compound as follows:- 1) The perElements 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

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


answered by: ANURANJAN SARSAM
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Answer #2

The weight. This empirical formula. can be calculated from the percentage Composition of the compound as follows:- 1) The perElements 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/molb.

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

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