[coffee]

Just came back to school.. and I realize I've forgotten so many things! ARGG! >_< So if anyone can, I would really appreciate it you could help and show/explain me how to do these problems!   


1) An element consists of 1.40% of an isotope with mass 203.973 amu, 24.10% of an isotope with mass 205.9745 amu, 22.10% of an isotope with mass 206.959 amu, and 52.40% of an isotope with mass 207.9766 amu. Calculate the average atomic mass and identify the elementy.

2) The element rhenium (Re) has two naturally occuring isotopes, 185Re and 187Re, with an average atomic mass of 186.207 amu. Rhenium is 62.60% 187Re, and the atomic mass of 187Re is 186.956 amu. Calculate the mass of 185Re.

3) A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound.

I have a few more problems, but hopefully after I solve these I might be able to solve the others, if not I might post more up. 

[BaO]

for questions 1 and 2 just use this formular :average atomic mass =( atomic mass1 x % amu +atomic mass 2 x % amu .)/ 100
if i'm not wrong the element number 1 has mentioned is Pb

for number 3, we have a reaction here : C_xH_yO_z + O₂ ->  CO2   + H2O   (not balanced)
                                                             10.68 mg                16.01mg  4.37mg
                                                             176.1 g/mol

# of moles of C_xH_yO_z  : 10.68 x10^(-3) g  x (1 mol/ 176.1g)  =  6.065 x 10^(-5) moles
------------------ CO2 : 16.01 x 10^ (-3) g  x (1mol / 44.0 g)  =  3.639 x 10 ^ (-4) moles
------------------ H2O : 4.37 x 10^(-3) g  x (1mol/ 18.0g)  =  2.428 x 10^(-4) moles
 
then we have a ratio for the above reaction  C_xH_yO_z : CO2  : H2O -> 1 : 6 :4

well, i hope up to here you can figure out its empirical formular .
 
one more hint: molecular formular = N x(empirical formular )   (where N=molar mass / empirical mass )

[AWK]

Bao's  method give us a molecular formula directly -  C6H8O6 (eg Vitamin C)

N in Bao's formula is called the greatest common divisor (in this case the GCD is 2, and empirical formula is C3H4O3).
Note - calculation of empirical mass can be omitted in this way.

[sdekivit]

just to give another way of calculation that leads to the same answer C₆H₈O₆:

16.01 mg CO₂ = 0.364 mmol CO₂ --> contains 0.364 mmol C: mass = 4.369 mg

4.37 mg H₂O = 0,243 mmol H₂O --> contains 0.486 mmol H: mass = 0.489 mg.

--> mass O: 10.68 - (4.369 + 0.489) = 5.822 mg --> thus 0.364 mmol O

Ratio C : H : O = 0.364 : 0.486 : 0.364 = 1 : 1.34 : 1 = 3 : 4 : 3

Empirical formula: C₃H₄O₃ --> molar mass = 88.062 g/mol

--> molar mass is 176.1, thus that is 176.1 / 88.062 = 2 times the empirical formula.

--> molecular formula: C₆H₈O₆.

[coffee]

Thanks everyone I got it now.