Question

When 0.22105g of it was burned in excess O2, 0.1422g of Bismuth (II) Oxide, 0.1880g of carbon dioxide, and .0.02750 g of water were formed. What is the empirical formula of this compound?

Answer 1

So far i have gotten as far as determining the moles of C, H, O, and Bi except my moles of O seem to be off not sure what I'm doing wrong...

Answer 2

6.10E-4 mole Bi
0.00427 mole C
0.00305 mole H
0.0138 mole O <---
i divided all by the smallest number and continued to multiply by the common coefficient but my number isn't the same as the back of the book answer: C7H5O4Bi

Answer 3

I thnk we need a balanced equation for this one to determine the moles of O2 needed during the reaction. Because 'excess of O2' throws us off.

Answer 4

well the initial statement is "When 0.22105g of \(\color{red}{\text{it}} \) " where that is what is being burned. So that is what I believe we are referring to here to find.
We have the skeletal equation for the most part, just the unknown is not here.
\( \text{Unknown} + \text{O}_2 \to \text{ BiO } + \text{CO}_2 + \text{H}_2 \text{O} \)
and we know the unknown is the limiting reagent here as well. We know the grams produced of each compound on the right along with that of the unknown. that should at least be enough to determine the mole ratios between the compounds...

Answer 5

yes @AccessDenied that is the equation i had written out as well!!

Answer 6

@iPwnBunnies ^^^

Answer 7

Okay, I think I get what we want to do.
First off, we have masses for almost everything. But how much O_2 do we need to react with our Unknown? Conservation of mass says we can find out.
,mass of Unknown + mass Oxygen gas = mass of Products combined
This becomes relevant later when we determine the amount of oxygen on the right, some of it is from the air and some of it may be in the compound.
We can determine the molar mass of each of our compounds:
O2 = 32 g/mol
BiO = 225 g/mol
CO2 = 44 g/mol
H2O = 18 g/mol
and use them to convert each mass we have into moles of each compound...
This seems correct so far? I will keep going.

Answer 8

We have...
0.13665 g O2 / (32 g/mol O2) = 0.004 270 mol O2
0.1422 g BiO / (225 g/mol BiO) = 0.000 632 mol BiO
0.1880 g CO2 / (44 g/mol CO2) = 0.004 273 mol CO2
0.0275 g H2O / (18 g/mol H2O) = 0.001 528 mol H2O

Answer 9

From here, we can associate moles of each element with the total moles of a compound.
There are 0.000 632 mol Bi within 0.000 632 mol BiO
There are 0.004 273 mol C within 0.004 273 mol CO2
&There are 2*0.001 528 mol H within 0.001 528 mol H2O

Answer 10

it is not a quiz

Answer 11

2* 0.004 270 mol O in .004 270 mol O2 on the left
.000 632 mol O per .000 632 mol BiO
.004 723*2 mol O per .004 723 mol CO2
.001 528 mol O per .000 528 mol H2O

So we could say our original compound has N moles of oxygen. We make the equation
N mol O + .0.004 270*2 mol O = .000 632 mol O + .004 723*2 mol O + .001 528 mol O
N mol O = .000 632 + .001 528 mol O
N = 0.002 160

.002 160 mol O
.000 632 mol Bi
.004 273 mol C
.001 528 mol H
Divide by the smallest value (0.000632)
3.41778 O
1 Br
6.76107 C
2.41772 H