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Chapter 10 Chemical Quantities

Chapter 10 Chemical Quantities

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Chapter 10 Chemical Quantities

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  1. Chapter 10Chemical Quantities

  2. Measuring Matter How much? How many? Chemistry is a quantitative science. How do you measure matter? You can measure the amount of something by count, by mass and by volume. Some units used for measuring indicate a specific number of items. (a pair, a dozen) Knowing how to relate the count, mass and volume to each other allows you to convert among the units.

  3. Measuring Matter What is the mass of 90 average-sized apples if 1 dozen of the apples has a mass of 2.0kg? 90 apples 1 dozen apples 2.0kg apples = 15kg apples 12 apples 1 dozen apples Matter is composed of atoms, molecules, and ions, which are difficult to measure individually. Chemists use a unit that is a specified number of representative particles, called a mole. A mole (mol) of a substance is 6.02 x 1023representativeparticles of a substance.

  4. Measuring Matter Avogadro’s number is the number of representative particles in a mole, 6.02 x 1023. The term representative particles refers to the species present in a substance: usually atoms, molecules or formula units. Representative particles for ionic compounds is the formula unit : CaCl2 , NaCl Representative particles for molecular compounds is the molecule: H2O , H2 Representative particles for most elements is the atom: Fe, Li

  5. Measuring Matter A mole of any substance contains Avogadro’s number of representative particles or 6.02 x 1023 representative particles. The relationship, 1 mole = 6.02 x 1023 representative particles, is the basis for a conversion factor to convert numbers of representative particles to moles. How many moles of Mg is 1.25 x 1023 atoms of Mg? 1.25 x 1023 atoms Mg (1 mol Mg / 6.02 x 1023 atoms Mg)

  6. Measuring Matter How many atoms are in 2.12 mol of propane (C6H8)? In the formula of a molecule of C3H8 , the subscripts show that propane is composed of 14 atoms: 3 atoms of C and 8 atoms of H. 2.12 mol C6H8 6.02 x 1023 molecules C6H8 11 atoms 1 mol C6H8 1 molecule of C6H8 1.40 x 1025 atoms

  7. Mass of a Mole The atomic mass of an element (mass of a single atom) is expressed in atomic mass units (amu) The atomic masses are relative values based on the mass of the most common isotope of carbon 12. The atomic mass of an element expressed in grams is the mass of a mole of the element. The mass of a mole of an element is its molar mass. Molar mass of C is 12.0 g. H – 1.0 g, S – 32.1g Molar mass is the atomic mass of an element rounded off to the first decimal place.

  8. Molar Mass If you were to compare 12.0g of C atoms with 16.0g of O atoms, you would find they contain the same number of atoms. The molar mass of any element contains 1 mole or 6.02 x 1023 atoms of that element. 12.0g of C is 1 mol of C atoms 1.0 g of H is 1 mol of H atoms Molar mass is the mass of 1 mole of atoms of any element.

  9. Mass of a Mole of a Compound To find the mass of a mole of a compound, you must know the formula of the compound. A molecule sulfur trioxide, SO3, is composed of one atom of sulfur and three atoms of oxygen. Calculate the mass of a molecule of SO3 by adding the atomic masses of the atoms making up the molecule. The atomic mass of Sulfur is 32.1g and the mass of three Oxygen atoms is 48.0g (3 x 16.0), so the molecular mass of SO3 is 80.1g (32.1 + 48.0) The molar mass of any compound is the mass of 1 mole of that compound.

  10. Mass of a Mole of a Compound 1 mole of SO3 has a mass of 80.1g and is the mass of 6.02 x 1023 molecules of SO3 To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound and then add the masses of the elements. The method for calculating molar mass applies to any compound, molecular or ionic.

  11. End of Section 10.1

  12. Mole/Mass Relationship You need 3.00 mol of NaCl. How do you measure this amount? What mass in grams is 3.00 mol of NaCl? 3.00 mol NaCl 58.5 g NaCl = 176g NaCl (use the molar mass) 1 mol NaCl When you measure 176g of NaCl on a balance, you are measuring 3.00 mol of NaCl. What is the mass of 9.45 mol of aluminum oxide? (Al2O3) 9.45 mol Al2O3 102.0g Al2O3 = 964 g Al2O3 1 mol Al2O3

  13. Mole/Mass Relationship How many moles of sodium sulfate (Na2SO4) is in 10 g of Na2SO4? 10.0 g Na2SO4 1 mol Na2SO4 = 7.04 x 10-2 mol Na2SO4 142.1 g Na2SO4 How many moles of iron(III) oxide are contained in 92.2 g of pure Fe2O3? 92.2 g Fe2O3 1 mol Fe2O3 = 0.578 mol Fe2O3 159.6 g Fe2O3

  14. Mole/Volume Relationship The volume of one mole of different solid and liquid substances are not the same. However, the volumes of moles of gases measured under standard condition are much more predictable. Avogadro’s hypothesis states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. If you buy a party balloon filled with helium and take it home on a cold day, you might notice that the balloon shrinks while it is outside. The volume of a gas varies with a change in temperature.

  15. Mole/Volume Relationship The volume of a gas also varies with a change in pressure. An increase in pressure causes the volume of the gas to decrease. Because of these variation due to temperature and pressure, the volume of a gas is usually measured at standard temperature and pressure. Standard temperature and pressure (STP) means a temperature of 0ºC and a pressure of 101.3 kPa (1atm) At STP, 1 mole or 6.02 x 1023 representative particles of any gas occupies 22.4L. 22.4 L is called the molar volume of gas.

  16. Mole/Volume Relationship If you have 0.375 mol of O2 gas, what volume at STP will this gas occupy? 0.375 mol O2 22.4L O2 = 8.40 L O2 1 mol O2 Determine the volume in liters of 0.60 mole of SO2 gas at STP. 0.60 mol SO2 22.4L SO2 = 13 L SO2 1 mol SO2 How many moles of H2 are in 0.200 L at STP? 0.200 L H2 1 mol H2 = 8.93 x 10-3 mol H2 22.4 L H2

  17. Molar Mass From Density Different gases have different densities. Usually the density of a gas is measured in grams per liter (g/L) The density of a gas at STP and the molar volume at STP can be used to calculate the molar mass of the gas. The density of a gaseous compound containing C and O is 1.964 g/L at STP. What is the molar mass of the compound? 1.964 g 22.4 L = 44.0 g/mol L 1 mol

  18. End of Section 10.2

  19. Percent Composition The relative amounts of the elements in a compound are expressed as the percent composition or the percent by mass of each element in the compound. The percent composition of a compound consists of a percent value for each different element in the compound. The percent composition of K2CrO4 is K = 40.3%, Cr = 26.8%, O = 32.9%. (They must total 100%) The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%.

  20. Percent Composition % mass of element = mass of element x 100% mass of compound When a 13.60 g sample of a compound containing only Mg and O is decomposed, 5.40g of O is obtained. What is the percent composition of this compound? % O = 5.40 g / 13.60g x 100% = 39.7% % Mg = 13.60 g – 5.40 g / 13.60g x 100% = 60.3%

  21. Percent Composition by Formula % mass = mass of element in 1 mol compound x 100% molar mass of compound Calculate the percent composition of propane C3H8 % C = 36.0 g / 44.0 g x 100% = 81.8% % H = 8.0 g / 44.0 g x 100% = 18.0%

  22. Percent Composition as a Conversion Factor How much C and H are contained in 82.0 g of propane? (C3H8) Calculate the percent composition of propane C3H8 % C = 36.0 g / 44.0 g x 100% = 81.8% % H = 8.0 g / 44.0 g x 100% = 18.0% In a 100 g sample of propane you would have 81.8 g of C and 18 g of O. (82.0 g propane)(81.8 g C / 100 g propane) = 67.1 g C (82.0 g propane)(18 g O / 100 g propane) = 15 g H

  23. Empirical Formulas The formula for some compounds show a basic ratio of elements. The percent composition is used to calculate the basic ratio of the elements contained in a compound. The basic ratio, called the empirical formula, gives the lowest whole-number ratio of the atoms of the elements in a compound. An empirical formula may or may not be the same as a molecular formula. The empirical formula of hydrogen peroxide is HO. The ratio of H to O is 1:1.

  24. Empirical Formulas The actual molecular formula of hydrogen peroxide is H2O2. The ratio of H to O is 1:1 In both cases the ratio of H to O is still the same 1:1 The molecular formula tells the actual number of each kind of atom present in a molecule of the compound.

  25. Calculating Empirical Formulas A compound is analyzed and found to contain 25.9% N and 74.1% O. What is the empirical formula of the compound? Convert each % to moles (25.9g N)(1 mol N / 14.0 g N) = 1.85 mol N (74.1g O)(1 mol O / 16 g O) = 4.63 mol O Next divide each molar quantity by the smaller number of moles to get a whole number ratio. 1.85 mol N/ 1.85 = 1 mol N 4.63 mol O / 1.85 = 2.50 mol O

  26. Calculating Empirical Formulas The result N1O2.5 still has a subscript that is not a whole number. If needed, to obtain the lowest whole number ratio, multiply each part by the smallest whole number that will convert the subscripts to whole numbers. 1 mol N x 2 = 2 mol N 2.5 mol O x 2 = 5 mol O The empirical formula is N2O5

  27. Molecular Formulas The molecular formula is either the same as its empirical formula, or it is a simple whole number multiple of its empirical formula. Once you have the empirical formula, you can determine the molecular formula, but you have to know the molar mass of the compound. Empirical formula for hydrogen peroxide is HO. Its empirical formula molar mass is 17.0 g/mol. The molar mass of hydrogen peroxide is 34.0 g.mol. 34.0 g/mol / 17.0 g/mol = 2 To obtain the molecular formula, multiply the subscripts in the empirical formula by 2. HO x 2 = H2O2

  28. Molecular Formulas • Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N. • Empirical formula molar mass is 30.9 g/mol • 60.0 g/mol / 30.0 g/mol = 2 • Molecular formula is CH4N x 2 = C2H8N2 • Find the molecular formula of ethylene glycol. The molar mass is 62.0 g/mol and the empirical formula is CH3O • 62.0 g/mol / 31.0 g/mol = 2 C2H6O2 • Which pair of molecules has the same empirical formula? • C2H4O2, C6H12O6 or b. NaCrO4, Na2Cr2O7 • a - If you multiply the first by 3 you get the second.

  29. End of Chapter 10 End of Chapter 7