mole calculation of chemistry
2017/06/25 05:09
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mole calculation of chemistry
"mole calculations of chemistry" 1.)What is the mass (in grams) of one molecule of CCl4?2.)How many chlorine atoms are present in 22.1 g of the compound?3.)A compound contains only carbon, hydrogen, and oxygen. Combustion of 48.64 g of the compound yields 71.30 g of CO2 and 29.19 g of... 顯示更多 "mole calculations of chemistry" 1.)What is the mass (in grams) of one molecule of CCl4? 2.)How many chlorine atoms are present in 22.1 g of the compound? 3.)A compound contains only carbon, hydrogen, and oxygen. Combustion of 48.64 g of the compound yields 71.30 g of CO2 and 29.19 g of H2O. The molar mass of the compound is 180.156 g/mol. 1. Calculate the grams of carbon (C) in 48.64 g of the compound. 2. Calculate the grams of hydrogen (H) in 48.64 g of the compound. 3. Calculate the grams of oxygen (O) in 48.64 g of the compound.
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1.)What is the mass (in grams) of one molecule of CCl4? Mass of 1 mole of CCl4 = 12 + 35.5x4 = 154 g No. of molecules in 1 mole of CCl4 = 6.02 x 1023 molecule Mass of 1 molecule of CCl4 = 154/(6.02 x 1023) = 2.56 x 10-22 g 2.) How many chlorine atoms are present in 22.1 g of the compound? No. of moles of CCl4 = 22.1/154 = 0.1435 mol Each mole of CCl4 contains 4 moles of Cl atoms. No. of moles of Cl atoms = 0.1435 x 4 = 0.574 mol Each mole of Cl atoms contains 6.02 x 1023 Cl atoms. No. of moles of Cl atoms = 0.574 x (6.02 x 1023) = 3.46 x 1023 atoms 3.) A compound contains only carbon, hydrogen, and oxygen. Combustion of 48.64 g of the compound yields 71.30 g of CO2 and 29.19 g of H2O. The molar mass of the compound is 180.156 g/mol. 1. Molar mass of C = 12 g/mol Molar mass of CO2 = 12 + 16x2 = 44 g/mol Mass fraction of C in CO2 = 12/44 Mass of C in 48.64 g of the compound = Mass of C in CO2 formed = 71.30 x (12/44) g = 19.45 g 2. Molar mass of H = 1 g/mol Molar mass of H2O = 1x2 + 16 = 18 g/mol Mass fraction of H in H2O = (1x2)/18 = 2/18 Mass of H in 48.64 g of the compound = Mass of H in H2O formed = 29.19 x (2/18) = 3.24 g 3. Mass of O in 48.64 g of the compound = 48.64 - (19.45 + 3.24) = 25.95 g [Mole ratio C : H : O = 19.45/12 : 3.24/1 : 25.95/16 = 1 : 2 : 1] [Empirical formula of the compound = CH2O] [Let (CH2O)n be the molecular formula of the compound.] [n(12 + 2x1 + 16) = 180.156] [n = 6] [Molecular formula = C6H12O6]
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mole calculation of chemistry
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發問:"mole calculations of chemistry" 1.)What is the mass (in grams) of one molecule of CCl4?2.)How many chlorine atoms are present in 22.1 g of the compound?3.)A compound contains only carbon, hydrogen, and oxygen. Combustion of 48.64 g of the compound yields 71.30 g of CO2 and 29.19 g of... 顯示更多 "mole calculations of chemistry" 1.)What is the mass (in grams) of one molecule of CCl4? 2.)How many chlorine atoms are present in 22.1 g of the compound? 3.)A compound contains only carbon, hydrogen, and oxygen. Combustion of 48.64 g of the compound yields 71.30 g of CO2 and 29.19 g of H2O. The molar mass of the compound is 180.156 g/mol. 1. Calculate the grams of carbon (C) in 48.64 g of the compound. 2. Calculate the grams of hydrogen (H) in 48.64 g of the compound. 3. Calculate the grams of oxygen (O) in 48.64 g of the compound.
最佳解答:
1.)What is the mass (in grams) of one molecule of CCl4? Mass of 1 mole of CCl4 = 12 + 35.5x4 = 154 g No. of molecules in 1 mole of CCl4 = 6.02 x 1023 molecule Mass of 1 molecule of CCl4 = 154/(6.02 x 1023) = 2.56 x 10-22 g 2.) How many chlorine atoms are present in 22.1 g of the compound? No. of moles of CCl4 = 22.1/154 = 0.1435 mol Each mole of CCl4 contains 4 moles of Cl atoms. No. of moles of Cl atoms = 0.1435 x 4 = 0.574 mol Each mole of Cl atoms contains 6.02 x 1023 Cl atoms. No. of moles of Cl atoms = 0.574 x (6.02 x 1023) = 3.46 x 1023 atoms 3.) A compound contains only carbon, hydrogen, and oxygen. Combustion of 48.64 g of the compound yields 71.30 g of CO2 and 29.19 g of H2O. The molar mass of the compound is 180.156 g/mol. 1. Molar mass of C = 12 g/mol Molar mass of CO2 = 12 + 16x2 = 44 g/mol Mass fraction of C in CO2 = 12/44 Mass of C in 48.64 g of the compound = Mass of C in CO2 formed = 71.30 x (12/44) g = 19.45 g 2. Molar mass of H = 1 g/mol Molar mass of H2O = 1x2 + 16 = 18 g/mol Mass fraction of H in H2O = (1x2)/18 = 2/18 Mass of H in 48.64 g of the compound = Mass of H in H2O formed = 29.19 x (2/18) = 3.24 g 3. Mass of O in 48.64 g of the compound = 48.64 - (19.45 + 3.24) = 25.95 g [Mole ratio C : H : O = 19.45/12 : 3.24/1 : 25.95/16 = 1 : 2 : 1] [Empirical formula of the compound = CH2O] [Let (CH2O)n be the molecular formula of the compound.] [n(12 + 2x1 + 16) = 180.156] [n = 6] [Molecular formula = C6H12O6]
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