What is the molecular formula of this compound
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# What is the molecular formula of this compound

[From: ] [author: ] [Date: 11-04-23] [Hit: ]
C=12.H=1.O=15.If100% = 58 40% = 23.26.67% = 3.......
What is the molecular formula of this compound?
An organic compound contains 54.55% C and 9.90% H. The molar mass is close to 90. What is the molecular formula?

Where I have gotten so far:

CxHyOz = 58g

C=12.001
H=1.0079
O=15.999

If 100% = 58
40% = 23.2
6.67% = 3.8744

Which leaves 53.33% (100% - 40% - 6.67%) = 30.9314

So that means there are 2 Carbon, 4 Hydrogen and 2 Oxygen.

How do I finish this off?

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Sorry, but I do not follow your method of calculating at all - where does: CxHyOz = 58g come from

I think that below is a fairly standardised method of working out these problems:
C = 54.55
H = 9.90
O = 100 - ( 54.55+9.90) = 35.55

Divide each by atomic mass:
C = 54.55 /12.011 = 4.542
H = 9.90/1.0079 = 9.822
O = 35.55/15.999 = 2.222

Divide by smallest:
C = 2.04
H = 4.42
O = 2.222/2.222 = 1

With respect to other answers - I do not think that you can round this to H = 4
If you round H to 4.5, then multiply through by 2 to remove decimal fraction
C = 4
H = 9
O = 2

Empirical formula = C4H9O2 molar mass 89.1135 g/mol - which is close to 90 as required
empirical and molecular formula = C4H9O2

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Assuming the remaining percentage is oxygen, we have the following. Assume a 100 gram sample so that each percentage can be expressed as a mass, and convert the masses to moles. Then divide each number of moles by the smallest number of moles to get integers. These become the subscripts in the empirical formula.
54.55% C
9.90% H
35.55% O

54.55 g C x (1 mol C / 12.01 g C = 4.542 / 2.22 = 2
9.90 g H x (1 mol H / 1.01 g H = 9.90 mol H / 2.22 = 4
35.55 g O x (1 mol O / 16.0 g O = 2.22 mol O / 2.22 = 1

The empirical formula is C2H4O, and its empirical molar mass is 44.05

If the actual molar mass is 90 g/mol, then there must be twice as many of each element, and so the molecular formula is C4H8O2.
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