Lesson 26 Empirical Formula Objectives: -         The student will determine an empirical, or simplest, formula from percentage composition data.   PA Science and Technology Standards: 3.4.10.A; 3.4.12.A PA Mathematics Standards: 2.2.11.A, 2.3.11.C, 2.4.11.E, 2.5.11.A

I. Formulas can be determined from composition. data. a I.          Formulas can be determined from composition data a.      When a new substance is discovered, the discoverers don’t usually know the formula b.      They do know the percentage composition c.      This can be used to determine the formula, by converting everything from percentages to moles. d.      These moles can then be divided by the smallest number that is determined, and this will give small, whole numbers which can be used to determine ratios.

e. Example. i. As part of a science fair project, Antonio is e.       Example       i.      As part of a science fair project, Antonio is analyzing the contents of fresh alkaline batteries. He has determined that one ingredient is a black powdery compound of 63% manganese and 37% oxygen. What is the compound’s formula?   List the percentages   Mn = 63% O = 37%

Determine the number of moles of each element present – do this by assuming you have 100 g of the compound in question.   1 mol Mn 63 g Mn X -------------- = 1.1467 --- = 1.1 mol Mn 54.94 g Mn   1 mol O 37 g O X ------------ = 2.3125 --- = 2.3 mol O 16 g O

Select the smallest number of moles, and divide all your number of moles by that one:     1.1  mol Mn --- = 1 mol Mn 1.1     2.3 mol O --- = 2.1 mol O 1.1   1.0  mol Mn and 2.1 (which is close to 2) mol O translates into:   MnO2 – the formula written in smallest whole numbers.

II. Recap: a. % to grams (assume 100g) b II.     Recap: a.       % to grams (assume 100g) b.      grams to moles c, divide by the smallest

III.       Special cases a.       Sometimes when you get to the last step, you don’t get numbers close to whole numbers. The following chart should help you with this:              i.      0.1-0.2 – round down.                         ii.      0.3 – multiply everything by 3                         iii.      0.4-0.6 – multiply everything by 2                         iv.      0.7 – multiply everything by 3                         v.      0.8-0.9 – round up b.      When completed, you should have moles close enough to whole numbers to make an empirical formula.

IV. Examples:

Empirical Formula Calculation The empirical formula is the formula with the lowest whole number ration between the elements. Example: the empirical formula of glucose which is C6H12O6 would be CH2O.   1. To calculate the empirical formula from the percent composition: 1.      Change the % signs to grams. 2.      Convert grams to moles 3.      Divide all the numbers of moles by the lowest number of moles. 4.      This gives you the ratio of the number of elements in the empirical formula.   Example: 11% Hydrogen and 89% Oxygen 11g H and 89 g O 11g H ( 1 mole/ 1.0079g) = 10.91378 moles   89 g O( 1 mole/ 15.999g) = 5.562848 moles   Oxygen has the smaller number of moles   H 10.91378/ 5.562848 = 1. 96 = about 2 O 5.562848/5.562848 = 1   Ratio is 2 H’s for every 1 O so the empirical formula is H2O

To calculate the empirical formula from experimental. data:. a To calculate the empirical formula from experimental data: a. convert the grams given into moles for each element and follow same procedure as above.

Empirical Formula Problems   Objectives: -         The student will convert percentage composition data into an empirical formula. -         The student will use supplied data to calculate a missing piece of data to complete the problem.   PA Mathematics Standards: 2.2.11.A, 2.4.11.E

Level 1 Determine the simplest formula for each compound listed below Level 1   Determine the simplest formula for each compound listed below. In the first eight problems, the percentage composition of the compound is given. In the last two, laboratory data for the compound are presented.   1. 80.0% carbon, 20.0 % hydrogen 2. 71.5% calcium, 28.5% oxygen 3. 82.2% nitrogen, 17.8% hydrogen 4. 85.7% carbon, 14.3% hydrogen 5. 6.6% aluminum, 93.4% iodine 6. 92.2% carbon, 7.8% hydrogen 7. 23.5% potassium, 76.5% iodine 8. 50.0 % sulfur, 50.0% oxygen 9. An oxide of arsenic contains 3.26 g of arsenic and 1.04 g of oxygen. What is the empirical formula for this oxide? 10. A sample of sodium oxide weighing 12.57 g contains 9.34 g of sodium. What is the empirical formula for this compound?

Level 2 Determine the simplest formula for each compound listed below Level 2   Determine the simplest formula for each compound listed below. In the first six problems, the percentage composition of the compound is shown. In the last four, laboratory data for the compound are given.   1. 63.5% silver, 8.2% nitrogen, 28.2% oxygen 2. 14.3% nitrogen, 4.1% hydrogen, 81.6% bromine 3. 24.7% potassium, 34.7% manganese, 40.5% oxygen 4. 56.6% potassium, 8.68% carbon, 34.7% oxygen 5. 9.92% carbon, 58.7% chlorine, 31.4% fluorine 6. 37.8% carbon, 6.4% hydrogen, 55.8% chlorine 7. 35.0% nitrogen, 5.0% hydrogen, 60.0% oxygen 8. 27.4% sodium, 1.2% hydrogen, 14.3% carbon, 57.1% oxygen 9. Analysis of a sample of a sulfur acid shows it to contain 0.17 g of hydrogen, 2.82 g of sulfur, and 5.67 g of oxygen. What is the simplest formula for this compound?

10. Analysis of a salt results in the following composition: 3.47 g of sodium, 2.12 g of nitrogen, and 7.27 g of oxygen. What is the empirical formula for this salt? 11. A barium salt is found to contain 21.93 g of barium, 5.12 g of sulfur, and 10.24 g of oxygen. What is the simplest formula of this compound? 12. An ore containing zinc, carbon, and oxygen, and weighing 485.35 g is analyzed and found to contain 46.59 g of carbon and 186.37 g of oxygen. What is the simplest formula for this compound?

Level 3 Determine the simplest formula for each compound listed below Level 3   Determine the simplest formula for each compound listed below. In the first six problems, the percentage composition of the compound is given. In the last four, laboratory data for the compound are given.   1. 26.6% potassium, 35.4% chromium, 38.0% oxygen 2. 74.0% carbon, 8.7% hydrogen, 17.3% nitrogen 3. 69.8% iron, 30.2% oxygen 4. 83.7% carbon, 16.3% hydrogen 5. 50.8% zinc, 16.0% phosphorus, 33.2% oxygen 6. 21.2% nitrogen, 6.1% hydrogen, 24.3% sulfur, 48.4% oxygen 7. Chemical analysis of a 10.000 g sample of oil of wintergreen shows that it consists of 6.32 g of carbon, 0.53 g of hydrogen, and 3.16 g of oxygen. What is the simplest formula for oil of wintergreen? 8. An acid is analyzed in the laboratory and the following results are obtained: 3.1% hydrogen, 31.6% phosphorus, 65.3% oxygen. What is the simplest formula for this acid?

9. Examination of 3. 2 x 10-2 g of an unknown white powder shows 9. Examination of 3.2 x 10-2 g of an unknown white powder shows that the powder consists of an unknown amount of nitrogen, 2.6 x 10-3 g of hydrogen, 6.7 x 10-3 of phosphorus, and 1.37 x 10-2 g of oxygen. What is the simplest formula for this compound? 10. A rock sample weighing 5.88 x 104 g is known to contain calcium, phosphorus, and oxygen. The amount of the first two elements in this rock is found to be 2.28 x 104 g and 1.18 x 104 g respectively. What is the formula for the compound in this rock sample?