percent composition & chemical formulas (empirical to molecular) chapter 10.3
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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
NameMolecular Formula
Multiple Molar mass Use or function
Formaldehyde CH2O 1 30Disinfectant,
biology preservative
Acetic Acid C2H4O2 2 60 Polymers, vinegar
Lactic Acid C3H6O3 3 90Sours milk,
product of exercise
Erythrose C3H6O3 4 120Forms during sugar
metabolism
Ribose C5H10O5 5 150Nuclei 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.