upon complete combustion 9.11g of carbohydrate gave 14.67g of carbon dioxide .another sample of 2.93g gave 1.93g of water .Find the molecular formula of the carbohydrate.
2006-12-03 21:47:53 · 2 個解答 · 發問者 ? 1 in 科學 ➔ 化學
9.11 g of the carbohydrate gave 14.67 g of CO2 :
Mass of C in CO2 = 14.67 x (12/44) = 4.00 g
Mass of C in the carbohydrate = 4.00 g
% by mass of C in the carbohydrate = 4.00/9.11 = 43.9%
2.93 g of the carbohydrate gave 1.93 of water :
Mass of H in H2O = 1.93 x (2/18) = 0.214 g
Mass of H in the carbohydrate = 0.214 g
% by mass of H in the carbohydrate = 0.214/2.93 = 7.3%
% by mass of O in the carbohydrate = (100-43.9-7.3)% = 48.8%
In the carbohydrate :
Mole ratio C : H : O
= (43.9/12) : (7.3/1) : (48.8/16)
= 3.66 : 7.3 : 3.05
= 1.2 : 2.4 : 1
= 6 : 12 : 5
Empirical formula = C6H12O5
It is not a carbohydrate because mole ratio H : O 2 : 1
More information is needed to calculate the molecular formula.
2006-12-04 12:10:53 補充:
It is not a carbohydrate because mole ratio H : O is not 2 : 1.
2006-12-04 07:09:21 · answer #1 · answered by Uncle Michael 7 · 0⤊ 0⤋
你不要死心不息啦!
你條題目只可以計到實驗式(empirical formula)。
沒有足夠數據去計分子式(molecular formula)。
2006-12-04 04:07:25 · answer #2 · answered by ? 7 · 0⤊ 0⤋