Empirical Formulas given quantities?
Give the empirical formula of each of the following compounds if a sample contains the following quantities of each element. (Type your answer using the format CO2 for CO2.)
(a) 0.0455 mol C, 0.039 mol H, and 0.013 mol O
(b) 46.84 g C, 11.79 g H, and 31.20 g O
(c) 54.5% C, 9.2% H, and 36.3% O by mass
if you could please explain how you got these answers, that would be great!
More Questions From This User See All
Answers & Comments
Verified answer
(a) we must divide by the smallest number
0.0455 / 0.013= 3.5 => C
0.039 / 0.013 = 3 => H
0.013 / 0.013 = 1 => O
we get C 3.5 H3 O
to have whole numbers we multiply by 2
C7H6O2 is the empirical formula
(b) we consider 100 g of this compound
Moles C = 46.84 g/ 12.011 g/mol = 3.900
Moles H = 11.79 g/ 1.008 g/mol = 11.70
Moles O = 31.20 g / 15.9994 g/mol = 1.950
we divide by the smallest number
C 2 H 6 O is the empirical formula
(c)
In the same way
Moles C = 4.538
Moles H = 9.127
Moles O = 2.269
we divide by the smallest number
C2 H 4 O is the empirical formula
You have to get the the ratio's of each element.
for a) work out the mass of C, H and O by using the formula:
Mass = Mol*atomic mass
Then use the emperical formulae of:
Mass/Relative atomic mass = ....
Then divide your answer by the smallest number to get your ratio. e.g. Mg = 0.6075g/ 24.3 = 0.025 and Br = 3.995g/79.9 = 0.050
0.025/0.025 = 1 and 0.050/0.025 = 2 so the resulting emperical formula is MgBr2.
b) Use the same formula as in part a. But you dont have to find the mass as it is already given.
c) and use the same formula aswell. Just substitute the percentages into where the mass is meant to be. so it is Percentage/relative atomic mass =.... / smallest number = the ratio. (This only works if all percentage values added together equal 100% as you turn the 54.5% of C into 54.5g of C is contained in a 100g compound and 9.2g of H and 36.3g of O.)