Question

02. (20 POINTS) An unknown substance is empirically determined to be 40 percent carbon by weight, 6.67 percent hydrogen, and 53.33 percent oxygen. Its molecular weight is roughly 55 g/mol. gml. Determine the molecular formula and its correct molecular weight. (a) List all the assumption you may use in the solution (b) The solution should be organized and neat (c) The final answer should be labeled ( draw a box around your answer) Question Max Points Points D1 25 3 20 04 20 5 15 Total 100 20 01. (25 POINTS) A room with volume of 500 m3, with fresh air entering at rate of 1000 m2/hr Suppose the air in the room is clean when it opens at 4 pm if a pollutant with concentration rate K-0.40 /hr is emitted from a smoke at constant rate 140 mg/hr starting 5 pm, what is the concentration at 6 pm?

Answer 1

Answer first question :

Assume we have 100 g of the compound.

Then we have 40.00 g of C, 6.67g of H, and 53.33g of O.

Moles of C = 40.00 g C × 1mol C/12.01g C = 3.331 mol C

Moles of H = 6.72 g H × 1 mol H/1.008g H = 6.67 mol H

Moles of O = 53.28 g O × 1 mol O/16.00g O = 3.330 mol H

Moles of C:Moles of H:Moles of O = 3.331:6.67:3.330=1.000:2.00:1 ≈ 1:2:1

The empirical formula is CH₂O.

The empirical formula is the simplest formula of a compound. The actual formula is an integral multiple of the empirical formula.

If the empirical formula is CH₂O, the actual formula is (CH₂O)n or CnH2nOn, where n = 1, 2, 3, … Our job is to determine the value of n.

The empirical formula mass of CH₂O is 30.03 u. The molecular mass of 180.18 u must be some multiple of this number.

for finding the molecular formula

So 55.0⋅g⋅mol−1=n×(12.01+2×1.008+16.00)g⋅mol−1.

n =1.896 ≈ 2

n=2 and the molecular formula is 2×(CH2O) = C2H4O2.

so molecular mass is 60 g/mol