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Chapter 7 Chemical Quantities

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.

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Chapter 7 Chemical Quantities

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  1. Chapter 7 Chemical Quantities 7.4 Mass Percent Composition and Empirical Formulas Learning Goal Given the formula of acompound, calculate the mass percentcomposition; from the mass percentcomposition, determine the empiricalformula of a compound. The odor of pears is due to propyl acetate, C5H10O2.

  2. Mass Percent Composition Given the mass of an element in a compound, we can calculate the mass percent composition ofthat element. The mass percent of an element in a compound isthe mass of an element divided by the total mass of the compound multiplied by 100%.

  3. Mass Percent Using Molar Mass 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%.

  4. Guide to Calculating Mass Percent

  5. Learning Check The odor of pears is due to the organic compound propyl acetate, C5H10O2. What is its mass percent composition? The odor of pears is due to propyl acetate, C5H10O2.

  6. Solution Propyl acetate has the formula C5H10O2; what is its mass percent composition? Step 1 Determine the total mass of each element in the formula. Given: C5H10O2 Need: mass percent composition

  7. Solution Propyl acetate has the formula C5H10O2; what is its mass percent composition? Step 1 Determine the total mass of each element in the formula.

  8. Solution Propyl acetate has the formula C5H10O2; what is its mass percent composition? Step 2 Divide the total mass of each element by the molar mass and multiply by 100%.

  9. Solution Propyl acetate has the formula C5H10O2; what is its mass percent composition? Step 2 Divide 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%

  10. Empirical Formulas 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 C5H10O5  5 = C1H2O1 = CH2O actual (molecular) empirical formula formula

  11. Empirical Formulas The molecular formula is the same or a multiple of the empirical formula.

  12. Learning Check A. What is the empirical formula for C4H8?(1) C2H4 (2) CH2 (3) CH B. What is the empirical formula for C8H14?(1) C4H7 (2) C6H12 (3) C8H14 C. Which is a possible molecular formulafor CH2O?(1) C4H4O4 (2) C2H4O2 (3) C3H6O3

  13. Solution A. What is the empirical formula for C4H8?(2) CH2 B. What is the empirical formula for C8H14?(1) C4H7 C. Which is a possible molecular formulafor CH2O?(2) C2H4O2 (3) C3H6O3

  14. Guide to Calculating Empirical Formula

  15. Learning Check A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula.

  16. Solution A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Step 1 Calculate the moles of each element.

  17. Solution A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Step 2 Divide by the smallest number of moles.

  18. Solution 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. Ni1.00Br2.00 = Ni1Br2, written as NiBr2

  19. Mass Percent The mass percent of CH4 is 74.9% carbon and 25.1% hydrogen.

  20. Converting Decimal Numbers to Whole Numbers When converting decimals to whole numbers, decimals greater than 0.1 or less than 0.9 should not be rounded off.

  21. Learning Check Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula.

  22. Solution Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. Step 1 Calculate 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.

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

  24. Solution 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.

  25. Solution 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)= C5H8O4

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