Right. Grams H + grams C = 1.24; I did that first and that checks out.
0.0918 mols C
0.137 mol H
0.0918/0.0918 = 1.00
0.137/0.0918 = 1.5
C1H1.5 = C2H3 with small whole numbers. I wonder if you rounded 1.5 to 2.0.
If you check out the equation, 1.24 g C2H3 will give you 4.04 g CO2 and I don't remember the H2O but I remember checking and that is ok too.
I agree with 0.0918 mols C and 0.137 mols H but not the formula.
0.137/0.0918 = about 1.5; is that C2H3?
And the balanced equation is
4C2H3 + 11O2 ==> 8CO2 + 6H2O
I've not heard of C2H3?
Okay, I have never heard ofC2H3 as well, so I had to take the number of moles that I calculated for carbon and hydrogen and see if they add up to the 1.24g of the hydrocarbon, which they did. So I go the formula CH2. We calculated the same numbers, but differ on formula. I'm not sure how???????
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I rounded 1.5 to 2, which is what you should do when they ask for the empirical formula. So, wouldn't the empirical formula be CH2 not C2H3?
I've never heard of either by the way.
Okay, never mind.
The ratio of C to H is 1 to 1.5, you have to multiply it by 2, and you get C2H3 to get whole numbers; 1.5 is to far from being a whole number.
The ratio of C to H is 1 to 1.5, you have to multiply it by 2, and you get C2H3 to get whole numbers; 1.5 is to far from being a whole number.
To just round up
Right. When rounding to a whole number one must be careful not to round too much. My rule of thumb is to look for whole numbers or multiply the ratios by 2,3,4,5 etc to try to get a whole number unless the value can be rounded to a whole number. Usually these problems come out so that anything, for example, between 1.2 and 1.8 PROBABLY can be multiplied by a whole number to get a whole number. So 1.1 or less I round to 1.0 and 1.9 or greater I round to 2.0. Values such as 0.667, 1.25, 1.33, 1.5. 1.667, 1.75 almost always can be multiplied by 4,3,2, etc to obtain a whole number.
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