Check question one and do others correctly by steps in general/college chemistry

Question

Check question one and do others correctly by steps in general/college chemistry 1
ula of an unknown compound with was determined in the following way: F 2. The form sample was analyzed by element, and found to consist of 54.55% carbon, 4318% fluorine by mass. Then irst, a 27% hydrogen, and a separate experiment was done in which a sample was vaporized 292-mL container in a water bath at 100°C. At this temperature, n a of 745 torr and was found to have a mass of 1.234 g. the gaseous sample had a eous sample had a PWhat is the empirical formula of this compound? What is the molecular formula?

Explanation / Answer

54.5% C

2.27 % H

43.18 % F

by mass

Assume a bsis of 100 g of sample

therefore

54.5 g C --> 54.5/12 = 4.5 mol C

2.27 g H --> 2.27/1 = 2.27 mol H

43.18 g F --> 43.18/19 = 2.27

Assume:

2.27 as basis:

1 mol o F

1 mol of H

2 mol of C

Empirical fomrula

C2HF

For MW

V = 292 ml

T = 100 C = 373 K

P = 745 torr = 745/760 = 0.98 atm

m = 1.234

Assume ideal gas

PV = nRT

PV = m/MW*RT

MW = m*RT/(PV) = 1.234*0.082*(373)/(0.082*0.292) = 1576.3 g/mol

Our Empriical formula was:

C2HF

Empirical mass = 12*2 + 1+ 19 = 44

then

1576.3 /44= 35.8 or about 39

then

C78H39F39

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