A 6.40g sample of a compound is burned to produce 8.37 g CO_2
, 2.75g H_2 O, 1.06g N_2 , and 1.23 g SO_2
Give your answer in the form C#H#N#O#S# where the number following the element’s symbol corresponds to the subscript in the formula. (Don’t include a 1 subscript explicitly)
What is the empirical formula of the compound?
2017-09-26 11:13 am
回答 (1)
2017-09-26 12:54 pm
Molar masses (in g/mol) :
H = 1.0, C = 12.0, N = 14.0, O = 32.0, S = 32.1
CO₂ = 44.0, H₂O = 18.0, N₂ = 28.0, SO₂ = 64.1
Mass of C in 6.40 g of the compound = Mass of C in CO₂ formed = (8.37 g) × (12.0/44.0) = 2.28 g
Mass of H in 6.40 g of the compound = Mass of H in H₂O formed = (2.75 g) × (1.0×2/18.9) = 0.306 g
Mass of N in 6.40 g of the compound = Mass of N₂ formed = 1.06 g
Mass of S in 6.40 g of the compound = Mass of S in SO₂ formed = (1.23 g) × (32.1/64.1) = 0.616 g
Mass of O in 6.40 g of the compound = [6.40 - (2.28 + 0.306 + 1.06 + 0.616)] g = 2.14 g
Molar ratio C : H : O : N : S
= 2.28/12.0 : 0.306/1.0 : 2.14/16 : 1.06/14 : 0.616/32.1
= 0.190 : 0.306 : 0.134 : 0.0757 : 0.0192
= 0.190/0.0192 : 0.306/0.0192 : 0.134/0.0192 : 0.0757/0.0192 : 0.0192/0.0192
= 9.90 : 15.9 : 6.98 : 3.94 : 1
≈ 10 : 16 : 7 : 4 : 1
Empirical formula = C₁₀H₁₆O₇N₄S
H = 1.0, C = 12.0, N = 14.0, O = 32.0, S = 32.1
CO₂ = 44.0, H₂O = 18.0, N₂ = 28.0, SO₂ = 64.1
Mass of C in 6.40 g of the compound = Mass of C in CO₂ formed = (8.37 g) × (12.0/44.0) = 2.28 g
Mass of H in 6.40 g of the compound = Mass of H in H₂O formed = (2.75 g) × (1.0×2/18.9) = 0.306 g
Mass of N in 6.40 g of the compound = Mass of N₂ formed = 1.06 g
Mass of S in 6.40 g of the compound = Mass of S in SO₂ formed = (1.23 g) × (32.1/64.1) = 0.616 g
Mass of O in 6.40 g of the compound = [6.40 - (2.28 + 0.306 + 1.06 + 0.616)] g = 2.14 g
Molar ratio C : H : O : N : S
= 2.28/12.0 : 0.306/1.0 : 2.14/16 : 1.06/14 : 0.616/32.1
= 0.190 : 0.306 : 0.134 : 0.0757 : 0.0192
= 0.190/0.0192 : 0.306/0.0192 : 0.134/0.0192 : 0.0757/0.0192 : 0.0192/0.0192
= 9.90 : 15.9 : 6.98 : 3.94 : 1
≈ 10 : 16 : 7 : 4 : 1
Empirical formula = C₁₀H₁₆O₇N₄S
收錄日期: 2021-04-24 00:41:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170926031329AADjyqr