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Chemistry Notebook: Percent Composition, Hydrates, Empirical and Molecular Formulas, Study notes of Chemistry

Martin Methodist CollegeChemistry

A detailed explanation of percent composition, hydrates, empirical formulas, and molecular formulas in chemistry. It includes examples and exercises to help understand the concepts. Students can use this document as study notes, summaries, or schemes and mind maps to prepare for exams.

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Name Date
Percent Composition, Hydrates, Empirical and Molecular Formulas!
Percent composition: the percentage by mass of each element in a compound.
• Percentage = (part / total)x100%
Hydrate: a chemical compound that contains chemically bound water molecules.
Empirical formula: simplest whole-number ratio of the atoms in the compound.
Molecular formula: whole-number multiple of the empirical formula.
• Shows the true composition.
• Molar mass of a compound = molar mass of the empirical formula x a whole number, n.
Part 1 (Review). Finding Percent composition. Divide the mass of the individual element within the compound by
the entire mass of the compound. Multiply by 100 to get the percent! (Part/whole *100%)
Ex. 1 mole of Fe(NO3)3 Contains 1 mole of iron, 3 moles of Nitrogen, and 9 moles of Oxygen.
Total molar mass = 55.845 + 3*14.007 + 9*15.999 = 241.85 g/mol
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 1 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀
𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚 (𝑁𝑁𝑂𝑂3)3
55.845𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚
241.85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗100% =23.1% 𝐹𝐹𝐹𝐹 𝑏𝑏𝑏𝑏 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚!
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 3 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑁𝑁 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀
𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚 (𝑁𝑁𝑂𝑂3)3
3∗14.007𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚
241.85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗100% =17.4% 𝑁𝑁 𝑏𝑏𝑏𝑏 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚!
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 9 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑂𝑂 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀
𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚 (𝑁𝑁𝑂𝑂3)3
9∗15.999𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚
241.85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗100% =59.5% 𝑂𝑂 𝑏𝑏𝑏𝑏 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚!
Now you try. Find the percent composition of Aluminum sulfate: Al2(SO4)3
Part 2 % composition
Empirical Formula
1. Convert % g.
2. Convert g mol : # g x 1𝑚𝑚𝑜𝑜𝑚𝑚
𝑚𝑚𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑚𝑚𝑀𝑀𝑀𝑀𝑀𝑀 𝑔𝑔= #𝑚𝑚𝑚𝑚𝑚𝑚
3. Compare these amounts in mol to find the simplest whole-number ratio among the elements.
(Divide each by the smallest number of moles. If you need to, double or even triple the answers to
make them into whole number ratios.) These ratios will be the subscripts.
4. Ex. Copper bromide is 28.5% Copper and 71.5% Bromide by mass. What is its empirical
formula?
Cu: 28.5 𝑔𝑔 𝐶𝐶𝐶𝐶 ∗1 𝑚𝑚𝑜𝑜𝑚𝑚 𝐶𝐶𝐶𝐶
63.546 𝑔𝑔 𝐶𝐶𝐶𝐶 = 0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 → 0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶
0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 = 1
Br: 71.5 𝑔𝑔 𝐵𝐵𝐵𝐵 ∗1 𝑚𝑚𝑜𝑜𝑚𝑚 𝐵𝐵𝑀𝑀
79.904 𝑔𝑔 𝐵𝐵𝑀𝑀 = 0.895 𝑚𝑚𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵 → 0.895 𝑚𝑚𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵
0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 = 1.997 ≈2
Ex. Aluminum oxide is 52.9% Aluminum and 47.1% Oxygen by mass. What is its empirical formula?
Al: 52.9 𝑔𝑔 𝐴𝐴𝑚𝑚 ∗1 𝑚𝑚𝑜𝑜𝑚𝑚 𝐴𝐴𝑚𝑚
26.982 𝑔𝑔 𝐴𝐴𝑚𝑚 = 1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 → 1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚
1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 = 1 ∗2 = 2
O: 47.1 𝑔𝑔 𝑂𝑂 ∗ 1 𝑚𝑚𝑜𝑜𝑚𝑚 𝑂𝑂
15.999 𝑔𝑔 𝑂𝑂= 2.94 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂→ 2.94 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂
1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 = 1.5 ∗2 = 3
Now you try. Find the empirical formula of a compound that is 46.6% Nitrogen and 53.4% Oxygen.
Fe
N
O
CuBr2
Al2O3
pf2

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Download Chemistry Notebook: Percent Composition, Hydrates, Empirical and Molecular Formulas and more Study notes Chemistry in PDF only on Docsity!

Name Date Percent Composition, Hydrates, Empirical and Molecular Formulas! Percent composition: the percentage by mass of each element in a compound.

  • Percentage = (part / total)x100% Hydrate: a chemical compound that contains chemically bound water molecules. Empirical formula : simplest whole-number ratio of the atoms in the compound. Molecular formula: whole-number multiple of the empirical formula.
  • Shows the true composition.
  • Molar mass of a compound = molar mass of the empirical formula x a whole number, n. Part 1 (Review). Finding Percent composition. Divide the mass of the individual element within the compound by the entire mass of the compound. Multiply by 100 to get the percent! (Part/whole *100%) Ex. 1 mole of Fe(NO 3 ) 3 Contains 1 mole of iron, 3 moles of Nitrogen, and 9 moles of Oxygen.

Total molar mass = 55.845 + 314.007 + 915.999 = 241.85 g/mol 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 1 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀 𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚(𝑁𝑁𝑂𝑂 3 ) 3 ^

55 .845𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 241 .85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗^ 100% = 23.1%^ 𝐹𝐹𝐹𝐹^ 𝑏𝑏𝑏𝑏^ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚! 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 3 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑁𝑁 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀 𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚(𝑁𝑁𝑂𝑂 3 ) 3 ^

3∗14.007𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 241 .85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗^ 100% = 17.4%^ 𝑁𝑁^ 𝑏𝑏𝑏𝑏^ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚! 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 9 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑂𝑂 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀 𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚(𝑁𝑁𝑂𝑂 3 ) 3 ^

9∗15.999𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 241 .85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗^ 100% = 59.5%^ 𝑂𝑂^ 𝑏𝑏𝑏𝑏^ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚! Now you try. Find the percent composition of Aluminum sulfate: Al 2 (SO 4 ) (^3)

Part 2 % compositionEmpirical Formula

  1. Convert %  g.
  2. Convert g  mol : # g x (^) 𝑚𝑚𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀1𝑚𝑚 𝑚𝑚𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀𝑀𝑀 𝑔𝑔 = #𝑚𝑚𝑚𝑚𝑚𝑚
  3. Compare these amounts in mol to find the simplest whole-number ratio among the elements. (Divide each by the smallest number of moles. If you need to, double or even triple the answers to make them into whole number ratios.) These ratios will be the subscripts.

4. Ex. Copper bromide is 28.5% Copper and 71.5% Bromide by mass. What is its empirical

formula?

Cu: 28.5 𝑔𝑔 𝐶𝐶𝐶𝐶 ∗ 631. 546 𝑚𝑚𝑜𝑜𝑚𝑚 𝑔𝑔𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 = 0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 → 0.448 0.448^ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚^ 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 1

Br: 71.5 𝑔𝑔 𝐵𝐵𝐵𝐵 ∗ 791.^904 𝑚𝑚𝑜𝑜𝑚𝑚 𝑔𝑔𝐵𝐵 𝑀𝑀𝐵𝐵𝑀𝑀 = 0.895 𝑚𝑚𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵 → 0.895 0.448^ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚^ 𝐵𝐵𝐵𝐵𝐶𝐶𝐶𝐶 = 1.997 ≈ 2

Ex. Aluminum oxide is 52.9% Aluminum and 47.1% Oxygen by mass. What is its empirical formula?

Al: 52.9 𝑔𝑔 𝐴𝐴𝑚𝑚 ∗ 261 .𝑚𝑚 982 𝑜𝑜𝑚𝑚 𝑔𝑔𝐴𝐴 𝑚𝑚𝐴𝐴𝑚𝑚 = 1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 → 1.96 1.96^ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚^ 𝐴𝐴𝐴𝐴𝑚𝑚𝑚𝑚 = 1 ∗ 2 = 2

O: 47.1 𝑔𝑔 𝑂𝑂 ∗ 151 .𝑚𝑚 999 𝑜𝑜𝑚𝑚 𝑔𝑔𝑂𝑂 𝑂𝑂 = 2.94 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂 → 1.962.94 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑂𝑂𝑚𝑚 = 1.5 ∗ 2 = 3

Now you try. Find the empirical formula of a compound that is 46.6% Nitrogen and 53.4% Oxygen.

Fe N O

CuBr 2

Al 2 O (^3)

Part 3 Hydrates!

Hydrates contain chemically bound water molecules. These waters can be removed by evaporation (heating). By subtracting the mass of the dehydrated compound from the mass of the hydrate, you can begin by finding the mass of the attached waters. If you know the percent composition of the hydrate, you can transform these into masses and start from this point as well.

1. Determine the mass of each the chemical compound and the bound/now freed waters.

2. Convert each mass into moles: # g x 𝑚𝑚𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀1𝑚𝑚 𝑚𝑚𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀𝑀𝑀 𝑔𝑔 = #𝑚𝑚𝑚𝑚𝑚𝑚

3. Determine the mole ratio (Divide each by the smallest number of moles. If you need to, double

or even triple the answers to make them into whole number ratios.) This ratios will tell you the

number of water molecules bound to the compound.

Ex. 100g of MgSO 4 •xH 2 O was heated and 48.8g of MgSO 4 remained: MgSO 4 ·xH 2 O (s) → MgSO 4 (s) + xH 2 O (g)

Mass of water: 100g M MgSO 4 •xH 2 O – 48g MgSO 4 = 52g H 2 O

MgSO 4 : 48.8𝑔𝑔 𝑀𝑀𝑔𝑔𝑀𝑀𝑂𝑂 4 ∗ 1201 𝑚𝑚. 366 𝑜𝑜𝑚𝑚^ 𝑀𝑀𝑔𝑔 𝑔𝑔𝑀𝑀𝑀𝑀𝑂𝑂𝑔𝑔𝑀𝑀𝑂𝑂^4 4

= 0.405 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑔𝑔𝑀𝑀𝑂𝑂 4 → 00 ..^405405 𝑚𝑚𝑚𝑚𝑜𝑜𝑜𝑜𝑚𝑚𝑚𝑚^ 𝑀𝑀𝑀𝑀𝑔𝑔𝑔𝑔𝑀𝑀𝑂𝑂𝑀𝑀𝑂𝑂^4

4

H 2 O: 51.2𝑔𝑔 𝐻𝐻 2 𝑂𝑂 ∗ 181 𝑚𝑚. 02 𝑜𝑜 𝑚𝑚𝑔𝑔^ 𝐻𝐻 𝐻𝐻^22 𝑂𝑂 𝑂𝑂 = 2.84 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻 2 𝑂𝑂 → 0. 4052.^84 𝑚𝑚^ 𝑚𝑚𝑜𝑜𝑜𝑜𝑚𝑚𝑚𝑚 𝑀𝑀^ 𝐻𝐻𝑔𝑔^2 𝑀𝑀𝑂𝑂𝑂𝑂 4 = 7 → 𝑥𝑥

Now you try: 10.00g of MgCO 3 •xH 2 O was heated and left 4.834 g of MgCO 3. What is the formula for the hydrate?

Part 4 EmpiricalMolecular Formulas Empirical formulas have only the simplest whole number ratio of elements, the molecular formula contains whole number multiples of these ratios.

1. Find the molar mass of the empirical formula.

2. Divide the experimentally determined (measured in a lab) mass of the molecule by the molar

mass of the empirical formula to determine what multiple of the empirical formula to use.

3. Multiply all subscripts by this multiple.

Ex. The empirical formula for a compound is P 2 O 5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.

Molar mass of the empirical formula P 2 O 5 : 2 × **molar mass of P = 61.94 g/mol

  • 5** × molar mass of O = 80.00 g/mol molar mass of P 2 O 5 = 141.94 g/mol

𝑛𝑛 =

n(empirical formula) = 2(P 2 O 5 ) = P 4 O 10

Now you try: A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula?

MgSO 4 • 7 H 2 O