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- 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. 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. 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. 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. Answer Total molar mass = 44.0 g/mol 36.0 g C = 81.8% 8.0 g H = 18.2%
- 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. 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. 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. 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. 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. 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. Answers Compound Empirical Formula HO 1. H 2O 2 2. CO2 CO2 3. N2H4 NH2 4. C6H12O6 5. HCN CH2O
- 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. 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. Answer Molar mass (efm) is 30.0 g 60.0 g divided by 30.0 g = 2 Answer: C2H8N2
- 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. Practice Problems 4) What is the molecular formula for each compound: a) CH2O: 90 g b) HgCl: 472.2 g c) C3H5O2: 146 g

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