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Understanding Percent Composition, Empirical and Molecular Formulas in Chemistry

This chapter delves into the concepts of percent composition, empirical formulas, and molecular formulas. It explains how to determine the formula mass of compounds like magnesium carbonate (MgCO3) and calculate their percentage composition. The empirical formula highlights the smallest whole number ratio of atoms, while the molecular formula provides the true number of atoms in a compound. It includes step-by-step calculations for determining empirical formulas and finding molecular formulas based on empirical data. Additionally, the chapter discusses hydrates and examples of ionic and molecular compounds.

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Understanding Percent Composition, Empirical and Molecular Formulas in Chemistry

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  1. Chapter 7 Section 3 and 4 Percent composition Empirical Formula Molecular Formula

  2. Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

  3. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

  4. Formulas • Empirical formula: the smallest whole number ratio of atoms in a compound. • molecular formula = (empirical formula)nn = whole # • empirical formula = CH • molecular formula = = (CH)6 C6H6 • Molecular formula: the true number of atoms of each element in the formula of a compound.

  5. Formulas(continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  6. Formulas(continued) Formulas for molecular compoundsMIGHT be empirical (lowest whole number ratio). Molecular: C6H12O6 H2O C12H22O11 Empirical: H2O CH2O C12H22O11

  7. Empirical Formula Determination • Use the % composition of each element out of 100 grams or given mass of each element in grams. • Convert the mass of each element to moles using molar mass. • Divide each value of moles by the smallest of the values to give the smallest whole # ratio of moles. • This should give the subscripts for the formula in the smallest whole numbers. If after dividing, the answer is not a whole number , multiply all by 2 or 3 to get a whole numbers.

  8. An unknown gas is found to be 24.0% carbon and 76.0% fluorine by mass. Determine the empirical formula of this gas. • 24.0g C x 1 mol = 2.00 mol 12.01 g C • 76.0 g F x 1 mol = 4.00 mol 19.00 g F 2.00/2.00 = 1 4.00/2.00 = 2 Multiply symbols by small whole # = C1F2or CF2 Work #2 on handout!

  9. Empirical Formula Determination A certain acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of the acid?

  10. Empirical Formula Determination(part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

  11. Empirical Formula Determination(part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 C3H5O2 Empirical formula: Work remaining problems on handout!

  12. Summary Steps: • Changes masses to moles • Divided each by the smallest mole # • Establish a whole number ratio

  13. Finding the Molecular FormulaDefinition – the true formula of a molecular compound. Sometimes called the “TRUE” formula! • Steps: • 1. Find the empirical formula. • 2. Divide molecular formula mass by empirical formula mass……MF/EF. • 3. Write molecular formula. The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  14. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  15. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

  16. Summary Steps for finding Molecular Formula: • Find the EF • Divide the given mass by the EF mass to create a ratio • Multiple the formula by the ratio # Work remaining problems on handout!

  17. Hydrates • Hydrates: ionic compounds with water incorporated into their structure • Ex. CuSO4•5H2O cupric sulfate pentahydrate C. Solve like an EF problem

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