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Percent comp_empirical formula_molecular formula

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Percent comp_empirical formula_molecular formula

1. 1. Terms to Know  Percent composition – relative amounts of each element in a compound  Empirical formula – lowest whole- number ratio of the atoms of an element in a compound
2. 2. An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound? 1. Calculate the total mass 2. Divide each given by the total mass and then multiply by 100% 3. Check your answer: The percentages should total 100%
3. 3. Answer  The total mass is 8.20 g + 5.40 g = 13.60 g  Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3%  Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7%  Check your answer: 60.3% + 39.7% = 100%
4. 4. Calculate the percent composition of propane (C3H8)  1. List the elements  2. Count the atoms  3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)  4. Express each element as a percentage of the total molar mass  5. Check your answer
5. 5. Answer  Total molar mass = 44.0 g/mol  36.0 g C = 81.8%  8.0 g H = 18.2%
6. 6. Calculate the mass of carbon in 52.0 g of propane (C3H8) Calculate the percent composition using the formula (See previous problem) 2. Determine 81.8% of 82.0 g Move decimal two places to the left (.818 x 82 g) 3. Answer = 67.1 g 1.
7. 7. Calculating Empirical Formulas  Microscopic – atoms  Macroscopic – moles of atoms  Lowest whole-number ratio may not be the same as the compound formula Example: The empirical formula of hydrogen peroxide (H2O2) is HO
8. 8. Empirical Formulas The first step is to find the mole-to-mole ratio of the elements in the compound  If the numbers are both whole numbers, these will be the subscripts of the elements in the formula  If the whole numbers are identical, substitute the number 1 Example: C2H2 and C8H8 have an empirical formula of CH  If either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts 
9. 9. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?    1. Assume 100 g of the compound, so that there are 25.9 g N and 74.1 g O 2. Convert to mole-to-mole ratio: Divide each by mass of one mole 25.9 g divided by 14.0 g = 1.85 mol N 74.1 g divided by 16.0 g = 4.63 mol O 3. Divide both molar quantities by the smaller number of moles
10. 10.      4. 1.85/1.85 = 1 mol N 4.63/1.85 = 2.5 mol O 5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number ) 2 x 1 mol N = 2 2 x 2.5 mol O = 5 Answer: The empirical formula is N2O5
11. 11. Determine the Empirical Formulas  1. H 2O 2  2. CO2  3. N2H4  4. C6H12O6  5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N?
12. 12. Answers  Compound Empirical Formula HO  1. H 2O 2  2. CO2 CO2  3. N2H4 NH2  4. C6H12O6  5. HCN CH2O
13. 13. Calculating Molecular Formulas  The molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula  The molecular formula may or may not be the same as the empirical formula
14. 14. Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N. 1. Using the empirical formula, calculate the empirical formula mass (efm) (Use the same procedure used to calculate molar mass.)  2. Divide the known molar mass by the efm  3. Multiply the formula subscripts by this value to get the molecular formula 
15. 15. Answer  Molar mass (efm) is 30.0 g  60.0 g divided by 30.0 g = 2  Answer: C2H8N2
16. 16. Practice Problems  1) What is the empirical formula of a compounds that is 25.9% nitrogen and 74.1% oxygen? 2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H. 3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.
17. 17. Practice Problems  4) What is the molecular formula for each compound: a) CH2O: 90 g b) HgCl: 472.2 g c) C3H5O2: 146 g