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Counting Large Quantities

Counting Large Quantities. Many chemical calculations require counting atoms and molecules It is difficult to do chemical calculations in terms of atoms or formula units Since atoms are so small, extremely large numbers are needed in calculations

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Counting Large Quantities

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  1. Counting Large Quantities • Many chemical calculations require counting atoms and molecules • It is difficult to do chemical calculations in terms of atoms or formula units • Since atoms are so small, extremely large numbers are needed in calculations • Need to use a special counting unit just as used for other items • A ream of paper • One dozen donuts • A pair of shoes

  2. The Mole • It is more convenient to use a special counting unit for such large quantities of particles • Mole: A unit that contains 6.022 х 1023 objects • It is used due to the extremely small size of atoms, molecules, and ions • 6.022x1023 particles in 1 mole • Called Avogadro’s Number • Periodic Table • The average atomic massinamu(one atom) • The weight of 1 mole of the element in grams • Avogadro’s number provides the connecting relationship between molar masses and atomic masses

  3. Calculating the Number of Molecules in a Mole • How many molecules of bromine are present in 0.045 mole of bromine gas? Avogadro’s number Given: 0.045 mol Br2 Need: molecules of Br2 Equality: Conversion factors: Set Up Problem:

  4. Subscripts State Moles of Elements • The subscripts in a chemical formula indicate the number of atoms of each element present in a compound • The subscripts in a chemical formula can also indicate the number of moles of atoms of each element present in one mole of a compound • i.e. In one molecule of glucose (C6H12O6) there are 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen

  5. Calculating the Moles of an Element in a Compound • How many moles of carbon atoms are present in 1.85 moles of glucose? subscript Plan: moles of glucose moles of C atoms Equality: (One) mol C6H12O6 = 6 mols C atoms Conversion Factors: 11.1 mol C atoms Set Up Problem:

  6. Molar Mass • The atomic mass of a carbon-12 atom is 12.00 amu • The atomic mass of one mole of carbon-12 atoms 12.00 g • One mole of any element is the amount of atoms (molecules or ions) that is equal to its atomic mass (in grams) • This mass contains 6.022 х 1023 particles of that element • Use the periodic table to obtain the molar mass of any element

  7. Molar Mass • When the number of grams (weighed out) of a substance equals the formula mass of that substance, Avogadro’s number of molecules of that substance are present

  8. Molar Mass of a Compound • Calculate the molar mass of iron (II) sulfate • Formula is FeSO4 • Calculate the molar mass of each element • Each element is multiplied by its respective subscript: (number of moles of each element) • The molar mass is calculated by the sum of the molar masses of each element Formula Subscript Moles of Element in Compound Moles of Compound

  9. Molar Mass of a Compound 1) Formula is FeSO4: The molar masses of iron, sulfur, and oxygen are 2) Multiply each molar mass by its subscript 3) Find the molar mass of the compound by adding the mass of each element

  10. Calculations Using Molar Mass • The three quantities most often calculated • Number of particles • Number of moles • Number of grams • Using molar mass as a conversion factor is one of the most useful in chemistry • Can be used for g to mole and mole to g conversions

  11. Relationship between Moles, Molar Mass and Avogadro’s number Avogadro’s Number Avogadro’s Number Avogadro’s Number Avogadro’s Number Particles of substance Particles of substance Moles of substance Moles of substance Moles of substance Moles of substance Molar Mass Grams of substance Moles of substance

  12. Converting Mass of a Compound to Moles • International Foods Coffee contains 3 mg of sodium chloride per cup of coffee. How many moles of sodium chloride are in each cup of coffee? 3 mg NaCl moles of NaCl = 0.003 g NaCl Equality: 1 mol NaCl = 58.44 g

  13. Converting Grams to Particles • Ethylene glycol (antifreeze) has the formula C2H6O2. How many molecules are present in a 3.86 × 10-20 g sample? Avog Number Molar mass Plan: convert g moles molecules of ethylene glycol Conversion Factor 1 Equality 1: Equality 2: Conversion Factor 2 375 molecules

  14. Percent Composition • Sometimes it’s useful to know the composition of a compound in terms of what percentage of the total is each element • Percent • “Parts per 100” • The number of specific items per a group of 100 items • 50% of $100 is $50 (50 items/100 total items)

  15. Percent Example • You have 4 oranges and 5 apples. What percent of the total is oranges? • In “parts per 100”

  16. Percent Composition • It is the percent by mass of each element in a compound • Can be determined • By its chemical formula • Molar masses of the elements that compose the compound • The percent of each element contributes to the mass of the compound

  17. Calculating Percent Composition Example • What is the percent composition of each element in NH4OH? Determine the contribution of each element Molar mass

  18. Empirical Formulas • The simplest ratio of elements in a compound • It uses the smallest possible whole number ratio of atoms present in a formula unit of a compound • If the percent composition is known, an empirical formula can be calculated

  19. Empirical Formulas • To Determine the empirical formula: • Calculate the moles of each element • Use molar mass (atomic mass) • Calculate the ratios of the elements to each other • Find the lowest whole number ratio • Divide each number of moles by the smallest number of moles present

  20. Empirical Formula: Converting Decimal Numbers to Whole Numbers • The subscripts in a formula are expressed as whole numbers, not as decimals • The resulting numbers from a calculation represent each element’s subscript • If the number(s) are NOT whole numbers, multiply each number by the same small integer (2, 3, 4, 5, or 6) until a whole number is obtained

  21. Relating Empirical and Molecular Formulas • n represents a whole number multiplier from 1 to as large as necessary • Calculate the empirical formula and the mass of the empirical formula • Divide the given molecular mass by the calculated empirical mass • Answer is a whole number multiplier

  22. Relating Empirical and Molecular Formulas • Multiply each subscript in the empirical formula by the whole number multiplier to get the molecular formula

  23. Calculate Empirical Formula from Percent Composition • Lactic acid has a molar mass of 90.08 g and has this percent composition: • 40.0% C, 6.71% H, 53.3% O • What is the empirical and molecular formula of lactic acid? • Assume a 100.0 g sample size • Convert percent numbers to grams

  24. Calculate Empirical Formula from Percent Composition • Convert mass of each element to moles • Divide each mole quantity by the smallest number of moles Empirical formula is CH2O The ratio of C to H to O is 1 to 2 to 1 Empirical formula mass = 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol

  25. Determination of the Molecular Formula • Obtain the value of n (whole number multiplier) • Multiply the empirical formula by the multiplier Molecular formula = n х empirical formula C3H6O3 Molecular formula = 3 (CH2O)

  26. Formulas for Compounds • Empirical Formula • Smallest possible set of subscript numbers • Smallest whole number ratio • All ionic compounds are given as empirical formulas • Molecular Formulas • The actual formulas of molecules • It shows all of the atoms present in a molecule • It may be the same as the EF or a whole- number multiple of its EF Molecular formula = n х Empirical formula

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