empirical & molecular formula
2010-05-31 8:58 pm
Compound X containing C,H,O only burned completely in air to form CO2 and H2O as the only products.2.43g of X gave 3.96 g of CO2 and 1.35g of H2O. Find the empirical formula of X. If relative molecular mass was 160, find also its molecular formula.
回答 (2)
2010-05-31 9:23 pm
✔ 最佳答案
Mass fraction of C in CO2 = 12/(12 + 16x 2) = 12/44Mass fraction of H in H2O = (1x2)/(1x2 + 16) = 2/18
In 2.43 g of X:
Mass of C = Mass of C in CO2 = 3.96 x(12/44) = 1.08 g
Mass of H = Mass of H in H2O = 1.35 x(2/18) = 0.15 g
Mass of O = 2.43 - (1.08 + 0.15) = 1.2 g
Mole ratio C : H : O
= (1.08/12) : (0.15/1) : (1.2/16)
= 0.09 : 0.15 : 0.075
= 1.2 : 2 : 1
= 6 : 10 : 5
Empirical formula of X = C6H10O5
Let (C6H10O5) be the molecular formula.
(12 x6 + 1x10 + 16x5)n = 160
162n = 160
n = 1
Molecular formula of X = C6H1oO5
參考: 老爺子
2010-05-31 11:11 pm
Mass fraction of C in CO2 = 12/(12 + 16x 2) = 12/44
Mass fraction of H in H2O = (1x2)/(1x2 + 16) = 2/18
In 2.43 g of X:
Mass of C = Mass of C in CO2 = 3.96 x(12/44) = 1.08 g
Mass of H = Mass of H in H2O = 1.35 x(2/18) = 0.15 g
Mass of O = 2.43 - (1.08 + 0.15) = 1.2 g
Mole ratio C : H : O
= (1.08/12) : (0.15/1) : (1.2/16)
= 0.09 : 0.15 : 0.075
= 1.2 : 2 : 1
= 6 : 10 : 5
Empirical formula of X = C6H10O5
Let (C6H10O5) be the molecular formula.
(12 x6 + 1x10 + 16x5)n = 160
162n = 160
n = 1
Molecular formula of X = C6H1oO5
Mass fraction of H in H2O = (1x2)/(1x2 + 16) = 2/18
In 2.43 g of X:
Mass of C = Mass of C in CO2 = 3.96 x(12/44) = 1.08 g
Mass of H = Mass of H in H2O = 1.35 x(2/18) = 0.15 g
Mass of O = 2.43 - (1.08 + 0.15) = 1.2 g
Mole ratio C : H : O
= (1.08/12) : (0.15/1) : (1.2/16)
= 0.09 : 0.15 : 0.075
= 1.2 : 2 : 1
= 6 : 10 : 5
Empirical formula of X = C6H10O5
Let (C6H10O5) be the molecular formula.
(12 x6 + 1x10 + 16x5)n = 160
162n = 160
n = 1
Molecular formula of X = C6H1oO5
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