Percent Composition & Chemical Formulas (empirical to molecular)

Chapter 10.3

Percent Composition • The percent by mass of each element in the

compound

How to find Percent Composition

• Percent composition can be determined in 2 ways.1. Determined using formulas

% mass of element = molar mass of element x 100

molar mass of compound

2. Found experimentally% mass of element = given mass of element x

100 given mass of compound

What is the % Composition ofPotassium Chromates?

• Determined using formulas • Potassium Chromate K2CrO4

% K = (2 x 39.1)/194.2 40.3%

% Cr = 52.0/194.2 26.8%% O = (4 x 16.0)/194.2

32.9%• Found experimentally• Upon analysis of a 5.00 g sample of a potassium chromate it was

found to contain 1.33g K, 1.77g Cr & 1.91g O. % K = 1.32/5.00

26.6%% Cr = 1.77/5.00 35.4%% O = 1.91/5.00

38.1%

Nitrogen (N)Key element in turf grass nutrition Promotes vigorous leaf and stem growth to improve the overall quality of the turf Essential component of the chlorophyll molecule which gives turf its dark green color Involved in regulating the uptake of other key elements

Phosphorous (P)Used in the formation and transfer of energy within the plant Influences early root development and growth Encourages plant establishment

Potassium (K)Used by the plant in large quantities, second only to nitrogen Key component in the formation of carbohydrates, or food for the plant Encourages rooting and wear tolerance Enhances drought and cold tolerance Key component in cell wall strength and resistance to disease

Fertilizers

What’s the right mix? What factors do you consider?

Empirical & Molecular Formulas • Empirical Formula

– A formula that gives the lowest whole-number ratio of elements in the compound.

– To get empirical formula from molecular, divide all subscripts by the greatest common multiple.

– Example: CH2O

• Molecular Formula – A formula that gives the actual ratio of elements in the

compound.– Molecular formula is a simple multiple of the empirical

formula– Example : C6H12O6

1.

4.

3.

5.

6.

2.

CH2O & CH2O

C2H4O2 & CH2O

C4H8O4 & CH2O

C3H6O3 & CH2O

What’s its formula? (CH2O)

C5H10O5 & CH2O

C6H12O6 & CH2O

Name Molecular Formula Multiple Molar mass Use or function

Formaldehyde CH2O 1 30 Disinfectant, biology preservative

Acetic Acid C2H4O2 2 60 Polymers, vinegar

Lactic Acid C3H6O3 3 90Sours milk,

product of exercise

Erythrose C3H6O3 4 120 Forms during sugar metabolism

Ribose C5H10O5 5 150 Nuclei acids RNA & vitamin B12

Glucose C6H12O6 6 180 Nutrient for energy

Empirical Formula CH2O

Going from % Comp to empirical formula1. Assume you have a 100g.

100 g means % directly coverts to grams

2. Convert elements to moles.3. Find a whole number ratio.

Divide each element by the smallest mole value to create a 1.

4. Write the empirical formula using the molar ratios as subscripts

79.85 %C = 79.85 g C 20.15 %H = 20.15 g H79.85 g C x 1 mol C = 6.65 mol C

12.0 g C20.15 g H x 1 mol H = 20.15 mol H

1.0 g H

C1H3 = CH3

_________ = 1 C 6.65 mol

________ = 3 H 6.65 mol

What if it doesn’t come out to a whole #1. Assume you have a 100g.

100 g means % directly coverts to grams

2. Convert elements to moles.3. Find a whole number ratio.

Divide each element by the smallest mole value to create a 1.Multiply all elements to generate a whole number 0.5 or ½ x 2 0.33 or ⅓ x 3 0.25 or ¼ x 4 0.2 or ⅕ x 5

4. Write the empirical formula using the molar ratios as subscripts

43.66 % P = 43.66 g P56.34% O = 56.34 g O 43.66 g P x 1 mol P = 1.41 mol P

31.0 g P56.34 g O x 1 mol O = 3.52 mol O

16.0 g OP2O5

_________ = 1 P 1.41 mol

_________ = 2.5 O 1.41 mol

X 2 = 2 P

X 2 = 5 O

Going from Empirical Formula to Molecular Formulas

1. Calculate the empirical formula mass

2. Divide the molar mass of the compound by the empirical formula mass.

3. Multiply this number by all subscripts in the empirical formula to get the molecular formula.

CH4N = 12.0 + (4 x 1.0) + 14.0 = 30.0 g

60.0g / 30.0 g = 2

CH4N x2 = C2H8N2

You can determine the molecular formula if you know empirical formula and molar mass.

? Find molecular formula with a molar mass of 60.0g and empirical formula of CH4N.