A compound contains 2.4 moles of carbon and 3.6 moles of oxygen. What is the percent composition by mass of each element in the compound?
回答 (5)
2016-06-19 12:18 pm
✔ 最佳答案
Molar mass of C = 12.01 g/molMass of C in the compound = (12.01 g/mol) × 2.4 = 28.8 g
Molar mass of O = 16.00 g/mol
Mass of O in the compound = (16.00 × 3.6) = 57.6 g
Total mass of the compound = (28.8 + 57.6) g = 86.4 g
% by mass of C in the compound = (28.8/86.4) × 100% = 33.3%
% by mass of O in the compound = 1 - 33.3% = 66.7%
2016-06-19 6:12 am
Weight of 3.6 moles of O at 16g/mol = 3.6 x 16 = 57.6g
Weight of 2.4 moles of C at 12g/mol = 2.4 x 12 = 28.8g
As you can see the oxygen weighs twice as much as the carbon hence the compound is CO2 or carbon dioxide.
My bad. Ignore that last part. I was misguided by what I saw as a flaw in your question.
Weight of 2.4 moles of C at 12g/mol = 2.4 x 12 = 28.8g
As you can see the oxygen weighs twice as much as the carbon hence the compound is CO2 or carbon dioxide.
My bad. Ignore that last part. I was misguided by what I saw as a flaw in your question.
2016-06-19 6:31 am
As a Medical Doctor I can confirm it is very harmful to the moles when you use fractional numbers of moles they do not deserve to be cut up like that
2016-06-29 11:41 am
carbon monoxide.
C=1.2
O=1.8
C=1.2
O=1.8
2016-06-19 9:30 am
Is this the complete question? There seems to be something missing:
If the compound contains 2.4 mol C and 3.6 mol O
C = 1
O = 1.5
And the empirical formula is C2O3 which is impossible.
Have you made a mistake in what you have submitted?
If the compound contains 2.4 mol C and 3.6 mol O
C = 1
O = 1.5
And the empirical formula is C2O3 which is impossible.
Have you made a mistake in what you have submitted?
收錄日期: 2021-04-18 15:06:42
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