Percent Composition - Formulas

Transcription

1 Percent Composition - Formulas This program demonstrates how to find percentage composition as well as empirical and molecular formulas from that data. Please read each section carefully. You might wish to take notes. Remember that all atomic masses are rounded to 3 significant figures. Percentage Composition Calculations To calculate the correct percentage composition of each element in some compound you must correctly follow several steps. It is not very difficult, once you get the hang of it. Please take notes on the steps to follow. Several problems follow this one for you to work. Lets look at a sample problem: magnesium chloride --> MgCl 2 step 1: calculate formula mass Mg: 1 x 24.3 = 24.3 Cl: 2 x 35.4 = formula mass = 95.1 step 2: divide each component mass by the formula mass and multiply by 100 Mg: 24.3/95.1 x 100 = 25.6% Cl: 70.8/95.1 x 100 = 74.4% step 3: make certain the percentages add up to 100 (+/- 0.1) = 100 This problem shows all the steps to follow. Be careful to use three significant figures at all times. Remember to first find the correct formula, then find formula mass, then calculate the percentage composition. The next two problems allow you some practice. Work these out on paper and then check your answer. Practice Problem #1 Find the percentage composition for each element in: iron (III) silicate iron (III) silicate --> Fe 2 (SiO 3 ) 3 Fe: 32.9% Si: 24.8% O: 42.4%

2 Remember that with significant figures your answer can be one above or below the final digit. Example: iron could have been reported as 32.8%, 32.9%, or even 33.0%. Work: Fe: 2 x 55.8 = /340. x 100 = 32.9% Si: 3 x 28.1 = /340. x 100 = 24.8% O: 9 x 16.0 = /340. x 100 = 42.4% (ok) Practice Problem #2 Find the percentage composition for each element in: tin (IV) arsenate. To see the correct answers and the method to solve the problem, please continue. tin (IV) arsenate --> Sn 3 (AsO 4 ) 4 Sn: 39.1% As: 32.8% O: 28.0% Work: Sn: 3 x 119 = /913 x 100 = 39.1% As: 4 x 74.9 = /913 x 100 = 32.8% O: 16 x 16.0 = /913 x 100 = 28.0% (ok) Empirical Formula Calculations Finding the empirical formula is somewhat the reverse of finding percentage composition. First you will be given the percentages of each element in a particular compound. Assuming that you will always be given a 100 gram sample we will convert these percentages to grams. The next step is to find the ratio of moles of each element and then calculate the simplest ratio of subscripts that maintains that ratio. It is much easier done than said! Take notes as we work through this sample problem. Lets start with this data: 36.5% Na, 25.4% S, and 38.1% O First, convert each percentage to grams: 36.5 g Na, 25.4 g S, 38.1 g O. Next, divide each by the grams/mole of that element: Na: 36.5 g = 1.58 mol Na 23.0 g/mol

3 S: 25.4 g = mol S 32.1 g/mol 0: 38.1 g = 2.38 mol O g/mol Now we can set up the ratio of moles of each element: Na S O To convert these decimal numbers into whole numbers and maintain the same ratio between them, just divide each by the smallest of the subscripts. Na S O Our final formula would look like: Na 2 SO 3 --> sodium sulfite Following will be two problems for you to work. Practice Problem #1 Given the following data, find the correct empirical formula: 49.0% C, 2.70% H, 48.2% Cl The correct formula would be: CH 3 Cl 2 To solve: C: 49.0g H: 2.70g Cl: 48.2g = 4.08 mol = 2.67 mol = 1.36 mol 12.0 g/mol 1.01 g/mol 35.5 g/mol

4 C H Cl gives the final formula: CH 3 Cl 2 Practice Problem #2 Given the following data find the empirical formula: N = 26.2%, H = 7.50%, Cl = 66.4% The correct formula would be: NH 4 Cl (ammonium chloride) The work: N: 26.2g H: 7.50g Cl: 66.4g = 1.87 mol = 7.50 mol = 1.87 mol 14.0 g/mol 1.01 g/mol 35.5 g/mol N H Cl > NH 4 Cl Molecular Formulas To find the molecular formula, first find the empirical formula for the data you have been given. Then compare the formula mass for your empirical formula with the formula mass for the molecular formula. You may have to multiply the subscripts in your empirical formula by some factor. Please take notes as we work through a sample problem. Sample Problem: Molecular Formula Calculation Given: 38.7% C, 9.70% H, 51.6% O and a molecular formula mass of 62.0 find the true molecular formula: First, find the empirical formula: --> CH 3 O

5 Next find the formula mass: C 1 x 12.0 = 12.0 H 3 x 1.01 = 3.03 O 1 x 16.0 = formula mass = 31.0 Now divide the molecular mass by this formula mass: 62.0/31.0 = 2 Multiply each subscript in your formula by this factor (2 here): Your final molecular formula: C 2 H 6 O 2 To check your answer find the formula mass of your final formula. It should be the same as the molecular formula mass. Practice Problem #1 Given the following data, find the correct molecular formula: 24.3% C, 4.1% H, 71.6% Cl molecular formula mass = 99.0 To see the correct answer and the method to solve the problem, please continue. The correct formula would be: C 2 H 4 Cl 2 The empirical formula would be CH 2 Cl with a formula mass of 49.5 Dividing 99.0 by 49.5 gives a factor of 2. Each subscript is doubled. Again, check the formula mass of your final answer to confirm your answer. Practice Problem #3 Given the following data, find the correct molecular formula: 54.6% C, 9.00% H, 36.4% O and a molecular mass of 176 The correct answer would be: C 8 H 16 O 4 The empirical formula would be: C 2 H 4 O with a formula mass of 44 The multiplication factor would be 4 (176/44.0)

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