solving stoichiometry problems

Solving Stoichiometry Problems

• Molecular weight of a substance

• Mole Conversions

• Isotopic abundances

• Empirical Formulas

• Combustion Analysis

• Empirical and Molecular Formulas

1Private Tutor David Alami – [email protected]

How to calculate the molecular weight of a substanceHere's how: multiply each element's atomic weight by how many atoms are present in the formula, then add the answers.

Example - Al2(SO4)3

There are: two atoms of aluminum and the atomic weight of Al is 26.98 amu.

three atoms of sulfur and the atomic weight of S is 32.06 amu.

twelve atoms of oxygen and the atomic weight of O is 16.00 amu.

First multiply:

2 x 26.98 = 53.96 total weight of all Al in formula

3 x 32.06 = 96.18 total weight of all S in formula

12 x 16.00 = 192.00 total weight of all O in formula

Then add: 53.96 + 96.18 + 192.00 = 342.14 amu.

This answer, 342.14 amu, represents the molecular weight of Al2(SO4)3

The molar mass of a substance is the molecular weight in grams.

All you need to do is calculate the molecular weight and stick the unit "g/mol" after the number andthat is the molar mass for the substance in question.


Mole Conversions: Given Moles, Convert to Grams

There are three steps to converting moles of a substance to grams:

• Determine how many moles are given in the problem.

• Calculate the molar mass of the substance.

• Multiply step one by step two.


Mole Conversions - Given Grams, Convert to Moles

• Determine how many grams are given in the problem.

• Calculate the molar mass of the substance.

• Divide step one by step two.


Mole Conversions Calculating the Mass of One Molecule

• Calculate the molar mass of the substance

• Divide it by Avogadro's Number

Example. Calculate the mass (in grams) of one molecule of CH3COOH

The molar mass of CH3COOH is 60.06 g/mol. The solution is:

60.06 g/mol ÷ 6.022 x 1023 mol¯1 = 9.973 x 10¯23 g


Calculate the isotopic abundances when given the averageatomic weight and the isotopic weights

(exact weight of isotope #1) (abundance of isotope #1) + (exactweight of isotope #2) (abundance of isotope #2) = average atomicweight of the element

Example: Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Givencopper's atomic weight of 63.546, what is the percent1) Write the following equation: (62.9296) (x) + (64.9278) (1 - x) = 63.546Once again, notice that 'x' and 'one minus x' add up to one. 2) Solve for x: x = 0.6915 (the decimal abundance for Cu-63)


Percent Composition

Percent composition is the percent by mass of each element present in a compound.1. figure out the molar mass from the formula.2. figure out the grams each atom contributes by multiplying the atomic weight bythe subscript.3. divide the answer for each atom by the molar mass and multiply by 100 to get a percentage.


Empirical Formula


Molecular Formula Empirical Formula

H2O H2O

CH3COOH CH2O

CH2O CH2O

C6H12O6 CH2O

The formula of a compound expressed as the smallest possible whole-number ratio of subscripts of theelements in the formula

Percent to mass

Mass to mole

Divide by small

Multiply 'til whole

Percent to mass: the assumption of 100 grams is purely

for convenience sake.

Example


A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular

weight of this compound is known to be approximately 140 g/mol. What is the empirical formula? What is the

molecular formula? 1) Percent to mass. Assume 100 grams of the substance is present, therefore its composition is:

carbon: 68.54 grams

hydrogen: 8.63 grams

oxygen: 22.83 grams

(2) Mass to moles. Divide each mass by the proper atomic weight.

carbon: 68.54 / 12.011 = 5.71 mol

hydrogen: 8.63 / 1.008 = 8.56 mol

oxygen: 22.83 / 16.00 = 1.43 mol

(3) Divide by small:

carbon: 5.71 ÷ 1.43 = 3.99

hydrogen: 8.56 ÷ 1.43 = 5.99

oxygen: 1.43 ÷ 1.43 = 1.00

(4) Multiply 'til whole. Not needed since all values came out whole.

The empirical formula of the compound is C4H6O

• 1) Determine the grams of each element present in the original compound. Carbon is always in CO2 in the ratio (12.011 g / 44.0098 g), hydrogen is always inH2O in the ratio (2.0158 g / 18.0152 g), etc. 2) Convert grams of each elment to the number of moles. You do this by dividingthe grams by the atomic weight of the element. 3) Divide each molar amount by the lowest value, seeking to modify the molaramounts into small, whole numbers.


Combustion Analysis

A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g ofH2O. What is the empirical formula of this compound? Solution:1) Determine the grams of carbon in 4.40 g CO2 and the grams of hydrogen in 2.70 g H2O. carbon: 4.40 g x (12.011 g / 44.0098 g) = 1.20083 g hydrogen: 2.70 g x (2.0158 g / 18.0152 g) = 0.3021482 g2) Convert grams of C and H to their respective amount of moles. carbon: 1.20083 g / 12.011 g/mol = 0.09998 molhydrogen: 0.3021482 g / 1.0079 g/mol = 0.2998 mol3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts intosmall, whole numbers. carbon: 0.09998 mol / 0.09998 mol = 1 hydrogen: 0.2998 mol / 0.09998 mol = 2.9986 = 3


Example

Empirical and Molecular Formulas

Combustion analysis can only determine the empirical formula of a compound; it cannotdetermine the molecular formula. However, other techniques can determine the molecularweight. Once we know this value, coupled with the empirical formulas, we can easily calculatewhat the molecular formula is.

Example. Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol. What is itsmolecular formula? Solution: (1) calculate the empirical formula, (2) compare "EFW" to molecular weight, (3) multiply empirical formula by proper scaling factor.

1) Calculate the empirical formula: • carbon: 49.98 g ÷ 12.011 g/mol = 4.16

hydrogen: 5.19 g ÷ 1.008 g/mol = 5.15nitrogen: 28.85 g ÷ 14.007 g/mol = 2.06oxygen: 16.48 g ÷ 15.999 g/mol = 1.03

• carbon: 4.16 ÷ 1.03 = 4.04 = 4hydrogen: 5.15 ÷ 1.03 = 5nitrogen: 2.06 ÷ 1.03 = 2oxygen: 1.03 ÷ 1.03 = 1

2) Empirical formula is C4H5N2O. The "empirical formula weight" is about 97.1, which gives a scaling factor of two. 3) The molecular formula is C8H10N4O2.


Homework problem

Hemoglobin is the protein that transports oxygen in mammals. Different species ofmammals have slightly different forms of hemoglobin. One form of hemoglobin is0.348% Fe by mass, and each hemoglobin molecule contains four iron atoms. Calculate the molar mass of this form of hemoglobin.

• SolutionAccording to the periodic table, the atomic mass for a Fe atom is 55.85 amu. Then4 Fe atoms have a mass of 4 x 55.85 amu = 223.4 amu. This represents 0.348% ofthe molecular mass of the hemoglobin. The decimal equivalent of 0.348% = 0.348/100 = 0.00348. If x = the molecular mass of the hemoglobin, then

223.4 amu = 0.00348x x = 66,700 amu

The molar mass = 64,195 g/mol


solving stoichiometry problems

Documents

total weight

averageatomic weight

atomic weight of s

atomic weight of al

atomic weight of o

elements atomic weight

givencoppers atomic

molar mass of ch3cooh