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© 2014 Pearson Education, Inc. Chapter 7 Lecture Basic Chemistry Fourth Edition Chapter 7 Chemical Quantities 7.4 Mass Percent Composition and Empirical Formulas Learning Goal Given the formula of a compound, calculate the mass percent composition; from the mass percent composition, determine the empirical formula of a compound. The odor of pears is due to propyl acetate, C 5 H 10 O 2.

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© 2014 Pearson Education, Inc. Given the mass of an element in a compound, we can calculate the mass percent composition of that element. The mass percent of an element in a compound is the mass of an element divided by the total mass of the compound multiplied by 100%. Mass Percent Composition

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© 2014 Pearson Education, Inc. Mass percent composition of a compound can be calculated using molar mass. The total mass of each element is divided by the molar mass of the compound, multiplied by 100%. Mass Percent Using Molar Mass

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© 2014 Pearson Education, Inc. Guide to Calculating Mass Percent

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© 2014 Pearson Education, Inc. The odor of pears is due to the organic compound propyl acetate, C 5 H 10 O 2. What is its mass percent composition? Learning Check The odor of pears is due to propyl acetate, C 5 H 10 O 2.

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© 2014 Pearson Education, Inc. Propyl acetate has the formula C 5 H 10 O 2 ; what is its mass percent composition? Step 1Determine the total mass of each element in the formula. Given: C 5 H 10 O 2 Need: mass percent composition Solution

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© 2014 Pearson Education, Inc. Propyl acetate has the formula C 5 H 10 O 2 ; what is its mass percent composition? Step 1Determine the total mass of each element in the formula. Solution

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© 2014 Pearson Education, Inc. Propyl acetate has the formula C 5 H 10 O 2 ; what is its mass percent composition? Step 2Divide the total mass of each element by the molar mass and multiply by 100%. Solution

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© 2014 Pearson Education, Inc. Propyl acetate has the formula C 5 H 10 O 2 ; what is its mass percent composition? Step 2Divide the total mass of each element by the molar mass and multiply by 100%. The total mass percent for all the elements should be 100%. 58.80% C + 9.870% H + 31.33% O = 100.00% Solution

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© 2014 Pearson Education, Inc. The empirical formula is the simplest whole-number ratio of the atoms is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio C 5 H 10 O 5 5 = C 1 H 2 O 1 = CH 2 O actual (molecular) empirical formula formula Empirical Formulas

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© 2014 Pearson Education, Inc. The molecular formula is the same or a multiple of the empirical formula. Empirical Formulas

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© 2014 Pearson Education, Inc. A.What is the empirical formula for C 4 H 8 ? (1) C 2 H 4 (2) CH 2 (3) CH B. What is the empirical formula for C 8 H 14 ? (1) C 4 H 7 (2) C 6 H 12 (3) C 8 H 14 C. Which is a possible molecular formula for CH 2 O? (1) C 4 H 4 O 4 (2) C 2 H 4 O 2 (3) C 3 H 6 O 3 Learning Check

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© 2014 Pearson Education, Inc. A.What is the empirical formula for C 4 H 8 ? (2) CH 2 B.What is the empirical formula for C 8 H 14 ? (1) C 4 H 7 C.Which is a possible molecular formula for CH 2 O? (2) C 2 H 4 O 2 (3) C 3 H 6 O 3 Solution

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© 2014 Pearson Education, Inc. Guide to Calculating Empirical Formula

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© 2014 Pearson Education, Inc. A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Learning Check

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© 2014 Pearson Education, Inc. A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Step 1Calculate the moles of each element. Solution

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© 2014 Pearson Education, Inc. A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Step 2Divide by the smallest number of moles. Solution

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© 2014 Pearson Education, Inc. A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Step 3Use the lowest whole-number ratio of moles as subscripts. Ni 1.00 Br 2.00 = Ni 1 Br 2, written as NiBr 2 Solution

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© 2014 Pearson Education, Inc. The mass percent of CH 4 is 74.9% carbon and 25.1% hydrogen. Mass Percent

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© 2014 Pearson Education, Inc. When converting decimals to whole numbers, decimals greater than 0.1 or less than 0.9 should not be rounded off. Converting Decimal Numbers to Whole Numbers

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© 2014 Pearson Education, Inc. Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. Learning Check

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© 2014 Pearson Education, Inc. Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. Step 1Calculate the moles of each element. A 100.0 g sample of aspirin contains 60.0 g C, 4.5 g H, and 35.5 g O. Solution

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© 2014 Pearson Education, Inc. Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. Step 1Calculate the moles of each element. Solution

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© 2014 Pearson Education, Inc. Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. Step 2Divide by the smallest number of moles. Solution

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© 2014 Pearson Education, Inc. Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. Step 3Use the lowest whole-number ratio of moles as subscripts. Multiply each of the subscripts by 4 to obtain a whole number. C (4×2.25) H (4×2.0) O (4 × 1.00) = C 5 H 8 O 4 Solution

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Copyright Sautter 2003. EMPIRICAL FORMULAE An empirical formula is the simplest formula for a compound. For example, H 2 O 2 can be reduced to a simpler.

Copyright Sautter 2003. EMPIRICAL FORMULAE An empirical formula is the simplest formula for a compound. For example, H 2 O 2 can be reduced to a simpler.

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