Chemistry Help. Please explain?

Question

Plz help on any of these 2 problems

Cinnamic acid contains only carbon, hydrogen and oxygen, and is found by analysis to be 73.0% C and 5.4% hydrogen. In a titration, 18.02 ml of 0.135M NaOH is found to neutralize 0.3602 g of cinnamic acid.

a. Find the empirical formula of this compound.

b. Find the molar mass of this compound.

c. Write the molecular formula for this compound.

Heating a 6.862 g sample of an ore containing a metal sulfide in excess oxygen produces 1053 ml of SO2 gas measured at 66 degrees C and 739 mmHg. Calculate the percentage by mass of sulfur in the ore.

Answer 1

**** a ****

73.0% C and 5.4% H with only C, H and O means 100% - 73.0% - 5.4% = %O = 21.6% O

if we assume 100 g of substance then we would have ....

73.0 g C = 73.0 g x ( 1 mole C / 12.0 g C ) = 6.0833 mole C

5.4 g H = 5.4 g x (1 mole H / 1.008g H) = 5.357 mole H

21.6 g O = 21.6 g x (1 mole O / 16.0 g O) = 1.35 mole O

divide by the smallest to get...

moles C = 6.0833 mole C / 1.35 = 4.5 moles C

moles H = 5.357 moles H / 1.35 = 4 moles H

moles O = 1.35 moles O / 1.35 = 1 mole O

multiple by 2 to get rid of the 4.5 and get whole numbers gives empirical formula to be

C9H8O2

**** b ****

assuming cinnamic acid to be a monoprotic carboxylic acid... (one C-O2H group)....

1 mole of NaOH would react with 1 mole of cinnamic acid...

18.02 ml of 0.135M NaOH has....

18.02 ml x (1 L / 1000 ml) x (0.135 moles NaOH / 1 L) = 0.002433 moles NaOH

so moles cinnamic acid would also be 0.002433 since moles NaOH = moles acid.

molar mass = grams / mole = 0.3602 g / 0.002433 mole = 148.1 g/mole

**** c ****

the formula mass of the empirical formula is

9 x 12.01 + 8 x 1.008 + 2 x 16.00 = 148.1 g

since the empirical formula mass = molar mass, 1 empirical unit = 1 molecule.

makes the molecular formula C9H8O2

*************************************

think of it like this...

MexSy + O2 ------> y SO2

if we calculate moles of SO2, we will know moles S.   We can then calculate mass of S and determine % mass of S in the ore....

assuming SO2 is ideal...

PV = nRT

n = PV/RT

R = .0821 Latm/moleK

T = 66+273.15 = 339.15 K

P = 739 mm Hg = 739 / 760 = .9724 atm

V = 1053 ml = 1.053 L

n = PV/RT = [(.9724 atm) x (1.053 L)] / [(.0821 Latm/moleK) x (339.15K)] = 0.03677 moles SO2

since 1 mole SO2 contains 1 mole S, we therefore have 0.03677 moles S

since atomic mass of S = 32.07 g/mole...

0.03677 moles S x (32.07 g S / mole S)= 1.179 g S

since all that sulfur came from the ore, the mass of sulfur in the ore is 1.179 g...

so the % S in the ore is 1.179 g / 6.862 g x 100% = 17.19%