Chem questions? molecular,Empirical formula, and molar mass?

1. A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.50 mg of the compound yields 15.74 mg CO2 and 4.30 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?

2. What mass of C2H3Cl3O2 would contain 3.8 g Cl?

____________ g C2H3Cl3O2


please show work so i know how you solved it, thank you

Answer:
1). So let's start by writing the reaction.

CxHyOz + O2 ---> CO2 + H2O

We know that all of the carbon on the product's side is from the unknown compound. We also know that all of the hydrogen on the product's side is from the unknown compound.

We have 15.74 mg CO2. Let's figure out the mass of carbon we have present:

15.74 mg CO2 x (1.000 g/ 1000. mg) = 0.01574 g CO2

0.01574 g CO2 x (1 mole/ 44.01 g) = 0.0003576 mole CO2

0.0003576 mole CO2 x (1 mole C/ 1 mole CO2)=0.0003576 mole C

0.0003576 mole C x (12.0 g/ 1 mole C) = 0.004291 g C

We have 4.30 mg H2O so let's figure out how much hydrogen we have:

4.30 mg H2O X (1.00 g/ 1000. mg) = 0.00430 g H2O

0.00430 g H2O x (1 mole/ 18.02 g)=0.000239 mole H2O

0.000239 mole H20 x (2 mole H/ 1 mole H20) = 0.000478 mole H

0.000478 mole H x (1.01 g/ 1 mole H)= 0.000483 g H

SO! we know we have 0.004291 g C and 0.000483 g H. We have 10.50 mg of the original sample. We have all the carbon and hydrogen accounted for so the rest of the weight is from oxygen:

10.50 mg x (1.000g / 1000. mg) = 0.01050 g total
0.01050 g - (0.004291 g C + 0.000483 g H) = 0.00573 g O

The formula is based on the mole ratio so we need to convert out masses to moles:

0.004291 g C x (1 mole/ 12.01 g C) = 0.000357 mole C
0.000483 g H x (1 mole/ 1.01 g H) = 0.000478 mole H
0.00573 g O x (1mole/ 16.00 g O) = 0.000358 mole O

The formula is based on whole number ratios. If we divide through the number of moles by the smallest number of moles we'll essentially base the ratios off of that. Remember any number divided by itself is one so when we divide through by the smallest number of moles the smallest number in the ratio is one, which is also the smallest whole number that can be used in a formula.

Essentially we have the same amount of carbon as we do oxygen so it really doesn't matter which we decide to divide by, but we'll just do carbon:

0.000357 mole C / 0.000357 mole C = 1 C
0.000478 mole H / 0.000357 mole C = 1.34 H
0.000358 mole O / 0.000357 mole C = 1 O

So the number of hydrogens is not a whole number but if we multiply all the numbers in the ratio by 3 we get whole numbers:

3 C : 4 H : 3 O

Therefore, the empirical formula is C3H4O3.

The molar mass of the compound is 176.1 g/mol.
C3H4O3 has a molar mass of 88.07 g/mol.

so if we divide the actual molar mass by this we can figure out the factor we have to multiply the subscripts by:

176.1/88.07 = 2

Therefore, the molecular formula is C6H8O6.

2).

3.8 g Cl x (1mole/ 35.45 g Cl) = 0.11 mole Cl

0.11 mole Cl x (1 mole C2H3Cl3O2/ 3 mole Cl) = 0.0367 mole C2H3Cl3O2

0.0367 mole C2H3Cl3O2 x (165.38 g/ 1 mole) = 6.070 g C2H3Cl3O2

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