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# A mass spectrometer

### Molar Mass

Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass

EXAMPLE 4

What is the molar mass of H2O?

SOLUTION

$\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol$

Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa.

EXAMPLE 5: COMBUSTION OF PROPANE

Propane ($$C_3H_8$$) burns in this reaction:

$C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2$

If 200 g of propane is burned, how many g of $$H_2O$$ is produced?

SOLUTION

Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of $$C_3H_8$$ to moles of $$C_3H_8$$ then from moles of $$C_3H_8$$ to moles of $$H_2O$$. Then convert from moles of $$H_2O$$ to grams of $$H_2O$$.

• Step 1:    200 g $$C_3H_8$$ is equal to 4.54 mol $$C_3H_8$$ .
• Step 2:    Since there is a ratio of 4:1 $$H_2O$$ to $$C_3H_8$$, for every 4.54 mol $$C_3H_8$$ there are 18.18 mol $$H_2O$$.
• Step 3:    Convert 18.18 mol $$H_2O$$ to g $$H_2O$$. 18.18 mol $$H_2O$$ is equal to 327.27 g $$H_2O$$.

### Variation in Stoichiometric Equations

Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis.

#### Density

Density ($$\rho$$) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used.

Volume x (Mass/Volume) = Mass

Mass x (Volume/Mass) = Volume

#### Percent Mass

Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule.

EXAMPLE 6

A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there?

SOLUTION

10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon

0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon