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the molecular weight of glucose

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

Molarity

Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions.

EXAMPLE 7

How much 5 M stock solution is needed to prepare 100 mL of 2 M solution?

SOLUTION

100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution.

These ratios of molarity, density, and mass percent are useful in complex examples ahead.

Determining Empirical Formulas

An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present.

EXAMPLE 8: COMBUSTION OF ORGANIC MOLECULES

1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO2 and 0.599 g of H2O. What is the empirical formula of the organic molecule?

SOLUTION

This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here.

\[ C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)\]

Since all the moles of C and H in CO2 and H2O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample.

0.0333mol CO2 (1mol C/ 1mol CO2) = 0.0333mol C in unknown

0.599g H2O (1mol H2O/ 18.01528g H2O)(2mol H/ 1mol H2O) = 0.0665 mol H in unknown

Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO2 and H2O. This will give you the number of moles from both the unknown organic molecule and the O2 so you must subtract the moles of oxygen transferred from the O2.

Moles of oxygen in CO2:

0.0333mol CO2 (2mol O/1mol CO2) = 0.0666 mol O

Moles of oxygen in H2O:

0.599g H2O (1mol H2O/18.01528 g H2O)(1mol O/1mol H2O) = 0.0332 mol O

Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O2 reactant.

0.333mol CO2(44.0098g CO2/ 1mol CO2) = 1.466g CO2

1.466g CO2 + 0.599g H2O – 1.000g unknown organic = 1.065g O2

Moles of oxygen in O2

1.065g O2(1mol O2/ 31.9988g O2)(2mol O/1mol O2) = 0.0666mol O

Moles of oxygen in unknown

(0.0666mol O + 0.0332 mol O) – 0.0666mol O = 0.0332 mol O

Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number.

(1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O

From this ratio, the empirical formula is calculated to be CH2O.