Question

http://prntscr.com/cifbc5

I know how to calculate empiricals by percent mass, but not this... :(

Answer 1

idea!

should I find the mole values for each set of '' grams?

Answer 2

yes
it is similar than the one that we did for the MSO4
here you get H2O and CO2
all the C from the CO2 is from the compound and all the H from the water it is also from the compound

Answer 3

hint says: "All of the carbon in carbon dioxide came from the original sample. So first convert the mass of CO2 to moles using the molar mass. Then determine the number of moles of C in that amount of CO2."

Answer 4

yes

Answer 5

the H2O you need it for the part B and remember you have 2 H in each H2O

Answer 6

alright

Answer 7

carbon dioxide = 12.01 (C) + 2*16.00 (O)

Answer 8

CO2=44g
H2O=18 g

Answer 9

Oh, thank you for that. :D

Answer 10

I got ~0.2340 mol for CO2

Answer 11

n=m/MM = 10.3/44 = 0.234 moles

Answer 12

4.20/18=0.233

0.233 x2= 0.466 moles of H

Answer 13

then it says: "If 7.00 g of the unknown compound contained 0.233 mol of C and 0.467 mol of H, how many moles of oxygen, O, were in the sample?"

Answer 14

all I know is that there are .233 mol CO2 and .233 mol H2O

Answer 15

.234 CO2 i mean

Answer 16

you have to calculate how many g of sample will have one mole of C

Answer 17

it says, "Because O2 is a reactant in combustion, we can't say that all the oxygen in CO2 and H2O came from the unknown. But we do know that the sum of all the elements in the unknown is equal to the total mass of the unknown. In other words,
(mass of C in sample)+(mass of H in sample)+(mass of O in sample)= 7.00 g
You determined the masses of carbon and hydrogen in Parts A and B. Solve for the mass of O in the sample."

Answer 18

m= n x MM
m= 0.233 mol of C x 12 =
m= 0.467 mol of H x 1=

Answer 19

I think I got it from here. is it okay to ask another Q post? (question)

Answer 20

also thank you for all the help :D

Answer 21

sure