If amount of h²o is 0.72 g nd CO² is 3.08 g. Then find the empirical formula.
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To begin this problem start with the amount of water formed. The molecular formula of water is H2O and the molecular weight (MW) is 18 g/mole. It is stated that you obtained 9 g of water and, in terms of moles of water, this is (9 g)/(18 g) = 0.5 mole.
For every mole of water (18 g), 16 g is from oxygen (MW = 16) and 2 are from hydrogen (MW = 1). In one-half mole, you will have 8 g from oxygen, 1 g from hydrogen (sum to 9 g so this makes sense). So, in 5.8 grams of hydrocarbon, 1 g of it is hydrogen, 4.8 g are from carbon. (Note, the problem says hydrocarbon so there can only be carbon and hydrogen in it and, thus, what is not hydrogen has to be carbon.)
Now, divide 4.8 g/12 g/mole to get the number of moles of carbon and you get 0.4 mole and for hydrogen 1g /1 g/mole = 1 mole. So the ratio of carbon to hydrogen is 0.4:1 or 1:2.5 (divide both by 0.4)
If this is an alkane (chain, no rings) the ratio of carbon to hydrogen is (n)/(2n+2) (the general empi rical formula for an C(n)H(2n+2)) and this has to equal 1/2.5:
n/(2n+2) = 1/2.5
2.5n = 2n+2
0.5n = 2
n = 4
and putting this back into C(n)H(2n+2) gives C4H10 - so this butane.
A little more effort to check things out. What if the hydrocarbon had a ring or was an mono alkene. The general formula for either is C(n)H(2n) and the carbon/hydrogen ratio is n/2n. Set this equal to the observed ratio (1/2.5) and you come up with something that cannot be true:
n/2n = 1/2.5
2.5n = 2n
which will never be true (at least in this universe/reality) so it can’t be an mono-alkene or mono-ring ring containing hydrocarbon.
What about 2 rings or an alkyne? The generalize formula for these is C(n)H(2n-2) and the carbon/hydrogen ratio = n/(2n-2). So, setting this equal to the observed value gives:
n/(2n-2) = 1/2.5
2.5n = 2n - 2
0.5n = -2
n = -4
This also is an impossible result as you can’t have a negative number of carbons (except, maybe if they are anti-matter carbons, lol)
For every mole of water (18 g), 16 g is from oxygen (MW = 16) and 2 are from hydrogen (MW = 1). In one-half mole, you will have 8 g from oxygen, 1 g from hydrogen (sum to 9 g so this makes sense). So, in 5.8 grams of hydrocarbon, 1 g of it is hydrogen, 4.8 g are from carbon. (Note, the problem says hydrocarbon so there can only be carbon and hydrogen in it and, thus, what is not hydrogen has to be carbon.)
Now, divide 4.8 g/12 g/mole to get the number of moles of carbon and you get 0.4 mole and for hydrogen 1g /1 g/mole = 1 mole. So the ratio of carbon to hydrogen is 0.4:1 or 1:2.5 (divide both by 0.4)
If this is an alkane (chain, no rings) the ratio of carbon to hydrogen is (n)/(2n+2) (the general empi rical formula for an C(n)H(2n+2)) and this has to equal 1/2.5:
n/(2n+2) = 1/2.5
2.5n = 2n+2
0.5n = 2
n = 4
and putting this back into C(n)H(2n+2) gives C4H10 - so this butane.
A little more effort to check things out. What if the hydrocarbon had a ring or was an mono alkene. The general formula for either is C(n)H(2n) and the carbon/hydrogen ratio is n/2n. Set this equal to the observed ratio (1/2.5) and you come up with something that cannot be true:
n/2n = 1/2.5
2.5n = 2n
which will never be true (at least in this universe/reality) so it can’t be an mono-alkene or mono-ring ring containing hydrocarbon.
What about 2 rings or an alkyne? The generalize formula for these is C(n)H(2n-2) and the carbon/hydrogen ratio = n/(2n-2). So, setting this equal to the observed value gives:
n/(2n-2) = 1/2.5
2.5n = 2n - 2
0.5n = -2
n = -4
This also is an impossible result as you can’t have a negative number of carbons (except, maybe if they are anti-matter carbons, lol)
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