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form 4 chem question.
1. 6.42g of a compound of carbon,hydrogen and oxygen gave on complete combustion 16.2g of cabon dioxide and 2.84g of water .Calculate the empirical formula of the original compound.2.a volatile liquid X is a compound of carbon, hydrogen and chlorine. 0.4 mole of X contains 9.6g of carbon 1.6g of hydrogen... 顯示更多 1. 6.42g of a compound of carbon,hydrogen and oxygen gave on complete combustion 16.2g of cabon dioxide and 2.84g of water .Calculate the empirical formula of the original compound. 2. a volatile liquid X is a compound of carbon, hydrogen and chlorine. 0.4 mole of X contains 9.6g of carbon 1.6g of hydrogen and 28.4 g of chlorine.Calculate the relative molecular mass and formula of compound X
最佳解答:
1. 16.2g CO2 x (1mol CO2/44.01g) x (1 mol C/ 1 mol CO2)=0.3680 mol C=4.421 g C 2.84g H2O x (1mol H2O/18.0158 g) x (2mol H/1mol H2O)=0.3153 mol H=0.3178 g H 6.42g compound-4.421g C-0.3178g H=1.6812g O=0.10507 mol O Since O has the least number of moles, divide the mole C and mole H by mole O, which gives you 3.5 mol C, 3 mol H and 1 mol O. Since you must have a whole number for all moles, multiply all moles by 2 gives you 7 mol C, 6 mol H, and 2 mol O. Therefore, the empirical formula is C7H6O2. 2. 9.6g C x (1mol C/12.01g)=0.799 mol C 1.6g H x (1mol H/1.0079g)=1.587 mol H 28.4 g Cl x (1mol Cl/35.4527g)=0.8036 mol Cl Since C has the least number of moles, divide all moles by the mole of C, which gives you 1 mol C, 1.98 mol H (which is rounded up to 2 mol H), and 1 mol Cl. The empirical formula is CH2Cl, with the mass of 12.01g C+ 1.0079g H x 2 +35.4527g Cl=49.4685g. Now we need to find the molecular mass and the molecular formula of the compound. Molar mass=gram/mole=(9.6g C+ 1.6g H+ 28.4g Cl)/0.4mole=99 grams/mole Our empirical formula CH2Cl gives 49.4685g/mol while the compound has 99g/mol, so if we double the empirical formula to C2H4Cl2, that is the molecular formula of the compound (and of course the molecular mass is 99g/mol).
其他解答:31C9A75CB3B14398
form 4 chem question.
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發問:1. 6.42g of a compound of carbon,hydrogen and oxygen gave on complete combustion 16.2g of cabon dioxide and 2.84g of water .Calculate the empirical formula of the original compound.2.a volatile liquid X is a compound of carbon, hydrogen and chlorine. 0.4 mole of X contains 9.6g of carbon 1.6g of hydrogen... 顯示更多 1. 6.42g of a compound of carbon,hydrogen and oxygen gave on complete combustion 16.2g of cabon dioxide and 2.84g of water .Calculate the empirical formula of the original compound. 2. a volatile liquid X is a compound of carbon, hydrogen and chlorine. 0.4 mole of X contains 9.6g of carbon 1.6g of hydrogen and 28.4 g of chlorine.Calculate the relative molecular mass and formula of compound X
最佳解答:
1. 16.2g CO2 x (1mol CO2/44.01g) x (1 mol C/ 1 mol CO2)=0.3680 mol C=4.421 g C 2.84g H2O x (1mol H2O/18.0158 g) x (2mol H/1mol H2O)=0.3153 mol H=0.3178 g H 6.42g compound-4.421g C-0.3178g H=1.6812g O=0.10507 mol O Since O has the least number of moles, divide the mole C and mole H by mole O, which gives you 3.5 mol C, 3 mol H and 1 mol O. Since you must have a whole number for all moles, multiply all moles by 2 gives you 7 mol C, 6 mol H, and 2 mol O. Therefore, the empirical formula is C7H6O2. 2. 9.6g C x (1mol C/12.01g)=0.799 mol C 1.6g H x (1mol H/1.0079g)=1.587 mol H 28.4 g Cl x (1mol Cl/35.4527g)=0.8036 mol Cl Since C has the least number of moles, divide all moles by the mole of C, which gives you 1 mol C, 1.98 mol H (which is rounded up to 2 mol H), and 1 mol Cl. The empirical formula is CH2Cl, with the mass of 12.01g C+ 1.0079g H x 2 +35.4527g Cl=49.4685g. Now we need to find the molecular mass and the molecular formula of the compound. Molar mass=gram/mole=(9.6g C+ 1.6g H+ 28.4g Cl)/0.4mole=99 grams/mole Our empirical formula CH2Cl gives 49.4685g/mol while the compound has 99g/mol, so if we double the empirical formula to C2H4Cl2, that is the molecular formula of the compound (and of course the molecular mass is 99g/mol).
其他解答:31C9A75CB3B14398
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