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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.
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 % composition Empirical Formula
Ex. Aluminum oxide is 52.9% Aluminum and 47.1% Oxygen by mass. What is its empirical formula?
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.
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
4
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 Empirical Molecular Formulas Empirical formulas have only the simplest whole number ratio of elements, the molecular formula contains whole number multiples of these ratios.
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
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