G An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of a 1.893- gram sample of the compound produced 5.545 grams of carbon dioxide and 1.388 grams of water. The 1.893-gram sample contained 7.002 x 10-3 moles. What is the molecular formula for this compound

C18H22O2

Explanation:

The following were obtained from the question:

Mass of the unknown compound = 1.893g

Mass of CO2 produced = 5.545g

Mass of H2O produced = 1.388g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Now let us obtain the various masses of C, H and O present in the unknown compound.

This is illustrated below:

Mass of C in the unknown compound = 12/44 x 5.545 = 1.512g

Mass of H in the unknown compound = 2/18 x 1.388 = 0.154g

Mass of O in the unknown compound = 1.893 — (1.512+0.154) = 0.227g

Now, let us calculate the empirical formula of the compound.

C = 1.512g

H = 0.154g

O = 0.227g

Divide by their molar Mass

C = 1.512/12 = 0.126

H = 0.154/1 = 0.154

O = 0.227/16 = 0.014

Now, divide by the smallest

C = 0.126/0.014 = 9

H = 0.154/0.014 =11

O = 0.014/0.014 = 1

The empirical formula is C9H11O

from the question,

Number of mole the unknown compound = 7.002x10^-3 mol

Mass of the unknown compound = 1.893g

Molar Mass of the unknown compound = Mass /number of mole = 1.893g/7.002x10^-3 mol = 270.35g/mol

Now, we can easily find the molecular formula of the unknown compound as follows

[C9H11O]n = 270.35

[(12x9) + (11x1) +16 ]n = 270.35

135n = 270.35

Divide both side by the coefficient of n i.e 135

n = 270.35/135 = 2

Therefore, the molecular formula is => [C9H11O]n

=> [C9H11O]2

=> C18H22O2

The molecular formula is C18H22O2

Explanation:

Step 1: Data given

Mass of the sample = 1.893 grams

Mass of CO2 produced = 5.545 grams

Mass of H2O produced = 1.388 grams

The 1.893-gram sample contained 7.002 * 10^-3 moles

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 5.545 grams / 44.01 g/mol

Moles CO2 = 0.1260 moles

In 1 mol CO2 we have 1 mol C

In 0.1260 moles CO2 we have 0.1260 moles C

Step 3: Calculate mass C

Mass C = 0.1260 moles * 12.01 g/mol

Mass C = 1.513 grams

Step 4: Calculate moles and H2O

Moles H2O = 1.388 grams / 18.02 g/mol

Moles H2O = 0.07703 moles

In 1 mol H2O we have 2 moles H

In 0.07703 moles H2O we have 2*0.07703 = 0.15406 moles H

Step 5: Calculate mass H

Mass H = 0.15406 moles * 1.01 g/mol

Mass H = 0.1556 grams

Step 6: Calculate mass O

Mass O = 1.893 - 1.513 - 0.1556

Mass O = 0.2244 grams

Step 7: Calculate moles O

Moles O = 0.2244 grams / 16.0 g/mol

Moles O = 0.01403 moles

Step 8: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.1260 moles/ 0.01403 moles = 9

H: 0.15406 moles / 0.01403 moles = 11

O: 0.01403 moles / 0.01403 moles = 1

The empirical formula is C9H11O

The molecular mass of this formula is 135.2 g/mol

Step 9: Calculate the molar mass of this compound

Molar mass = 1.893 grams / 0.007002 moles

Molar mass = 270.4 g/mol

Step 9: Calculate the molecular formula

We have to multiply the empirical formula by n

n = 270.4 / 135.2  = 2

The molecular formula is 2*(C9H11O) = C18H22O2

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !