a compound containing Fe, C, H and O A compound containing Fe, C, H and O was de

Question

a compound containing Fe, C, H and O A compound containing Fe, C, H and O was determined by combustion analysis to have 51.04% C and 5.98% H in a separate experiment to determine the mass % of Fe, 0.3355 g of the compound yielded 0.0758 g of Fe _2 O what is the empirical formula of the compound ?Questions to consider in completing this problem:what 2 quantities should you determine first in order to find the empirical formula ? Which of the quantities in (a) above will be the same Forall both Fe _2 O _3 and the unknown compound how will you determine the mass percentage of O in the compound? F + O _2 rightarrow H _2 O + CO _2 100-57.02 = 42.98 100g sample

Explanation / Answer

0.3355g of the compound yields 0.0758g of Fe2O3

Now, mol.wt of Fe2O3=159.69 g/mol and atomic wt of Fe= 55.86

Also 1 mole of Fe2O3 contains 2 moles of Fe

i.e 159.69 g of Fe2O3 contains 2 x 55.84 = 111.68 g of Fe

So 0.0758g of Fe2O3 contains (111.68 x 0.0758)/159.69 = 0.053 g of Fe

Hence 0.3355g of the compound contains 0.053 g of Fe

% Fe = 0.053/0.3355 x 100 = 15.80 %

% C = 51.04, % H = 5.98 (given)

Thus % O = 100- (51.04 + 5.98 + 15.80) = 27.18 %

Now Assuming 100g of compound is present,

C = 51.04 g, H = 5.98 g, Fe = 15.80 g and O=27.18 g

Calculate the respective number of moles for each element by dividing with the corresponding atomic weights

C : 51.04/12 = 4.2533 moles

H : 5.98/1 = 5.98 moles

Fe: 15.80/55.84 = 0.2829 moles

O: 27.18/16 = 1.6988 moles

Now divide by the lowest value (0.2829 in this case) to get whole number ratios

C : 4.2533/0.2829 = 15.0

H: 5.98/0.2829 = 21.1

Fe: 0.2829/0.2829 = 1.0

O: 1.6988/0.2829 = 6.0

Empirical Formula C15H21O6Fe

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