Chapter 10 Chemical Quantities. Molar Mass  MOLAR MASS—# of grams of an element in one mole of that element. = atomic mass of an element with the units.

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Chapter 10 Chemical Quantities
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Presentation transcript:

Chapter 10 Chemical Quantities

Molar Mass  MOLAR MASS—# of grams of an element in one mole of that element. = atomic mass of an element with the units of g/mol Get it from the periodic table Get it from the periodic table

Elements Compounds ADD the molar mass of every atom present in the compound Find the molar mass on the periodic table round to 0.1 g/mol Molar Mass

M OLAR M ASS P RACTICE

If you know the MOLAR MASS … You can convert between grams and moles! You can convert between grams and moles! 1 Mole Mass Volume of gas at STP Representative Particles Molar mass (g/mol)

Mole ↔ Mass Examples  How many moles in 51.0 g of ?  How many moles in 51.0 g of Na 3 PO 4 ? Find Molar Mass: (23.0 x 3) (16.0 x 4) = 164 g/mol   51.0 g Na 3 PO 4 x 1 mole Na 3 PO 4 = mol Na 3 PO g Na 3 PO 4  What is the mass of 0.70 mol of ?  What is the mass of 0.70 mol of N H 4 Cl? Find Molar Mass: (1.0 x 4) = 53.5g/mol   0.70 mol NH 4 Cl x 53.5 g NH 4 Cl = g NH 4 Cl 1 mol NH 4 Cl

If you know Avogadro’s number … You can convert between particles and moles! You can convert between particles and moles! 1 Mole Mass Volume of gas at STP Representative Particles Avogadro’s Number 6.02 x Molar mass (g/mol) Atoms – elements Formula Units - ionic Molecules - covalent A mole = Avogadro’s # = 6.02x10 23 representative particles By COUNT:

By COUNT: A A A A A vvvv oooo gggg aaaa dddd rrrr oooo '''' ssss N N N N uuuu mmmm bbbb eeee rrrrHow big is 6.02x1023?  If you had Avogadro's number of un-popped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. Avogadro's number Avogadro's number  If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. countone molecountone mole  A mole of coke cans would cover the surface of the earth to a depth of over 200 miles. mole

Measuring Matter  Some units of measurement indicate specific numbers. A pair means 2 A pair means 2 A dozen means 12 A dozen means 12  Knowing how count, mass, & volume relate allows you to convert between them.  If 1 dozen apples = 12 apples, and 1 dozen apples has a mass of 2.0 kg, What is the mass of 90 apples? 90/12 = 7.5 dozen 7.5 dozen x 2.0 kg 90/12 = 7.5 dozen 7.5 dozen x 2.0 kg = 15 kg = 15 kg

What is a Mole?  When dealing with tiny particles (atoms, ions, compounds), the sample size is usually very large.  Counting is not practical.  Just as a dozen eggs represents 12 eggs, a mole (mol) of a substance represents 6.02 x representative particles of that substance.  The number 6.02 x is known as Avogadro’s number

The Mole  A representative particle tells the type of particle: atoms, molecules, or formula units  1 mole of nitrogen gas (N 2 ) contains 6.02 x molecules. Because N 2 is a molecule Because N 2 is a molecule  1 mole of calcium fluoride (CaF 2 ) contains 6.02 x formula units Because CaF 2 is an ionic compound, and the representative particle is the formula unit. Because CaF 2 is an ionic compound, and the representative particle is the formula unit.  How many atoms of He are in 1 mole?

Moles  Determining the number of atoms in a mole of a compound: How many moles are in a representative particle (formula unit or molecule) of the substance? How many moles are in a representative particle (formula unit or molecule) of the substance? This can be determined from the formula. This can be determined from the formula.  Example: each molecule of CO 2 contains 3 atoms: 1 carbon atom and 2 oxygen atoms. A mole of CO 2 contains 1 mole of carbon atoms and 2 moles of oxygen atoms. A mole of CO 2 contains 1 mole of carbon atoms and 2 moles of oxygen atoms.

The Mass of a Mole  When dealing with atoms, it is often easier to work with mass.  Gram atomic mass (gam)—the atomic mass of an element expressed in grams. The atomic mass of carbon is 12 amu The atomic mass of carbon is 12 amu The gam of carbon is 12 g. The gam of carbon is 12 g.  The gam is the mass of 1 mole of atoms of any element.

The Mass of a Mole of a Compound  To determine the mass of a mole of a compound you need to know the formula and the gam of each atom in the compound. Add the masses of each atom to get the mass of the compound Add the masses of each atom to get the mass of the compound  Sulfur trioxide (SO 3 ) 1 mole of S (32.1g) & 3 moles of O (16.0 g each) 1 mole of S (32.1g) & 3 moles of O (16.0 g each) 32.1g g g g = 80.1 g 32.1g g g g = 80.1 g  Gram molecular mass (gmm)—the mass of 1 mole of a molecular compound  Gram formula mass (gfm)—the mass of 1 mole of an ionic compound

Sec. 2 Mole-Mass-Volume  We learned about gam, gmm, & gfm last time. We can use 1 broad term to tell the mass of a substance.  Molar Mass—the mass (in grams) of 1 mole of a substance. Why do we have the 3 terms then? Sometimes the term molar mass in unclear. What is the molar mass of oxygen? Do you mean oxygen gas (O 2 )? Then the molar mass is 32.0g (2 x 16.0g). Or do you mean oxygen atoms (O)? Then the molar mass is 16.0 g. Why do we have the 3 terms then? Sometimes the term molar mass in unclear. What is the molar mass of oxygen? Do you mean oxygen gas (O 2 )? Then the molar mass is 32.0g (2 x 16.0g). Or do you mean oxygen atoms (O)? Then the molar mass is 16.0 g.

The Volume of a Mole of Gas  The volumes of 1 mole of different solid and liquid substances are not the same.  The volumes of 1 mole of different gases are the same under the same conditions.  To keep things under the same conditions, gases are measured at standard temperature and pressure (STP)  Standard temp is 0°C  Standard pressure is 101.3kPa (or 1 atm)  At STP, 1 mole of any gas has a volume of 22.4 L

Converting Between moles, particles, mass, and volume Mole Mass Volume of gas at STP Representative Particles Molar mass 1 mol 1 mol. molar mass 1 mol 22.4 L 1 mol 6.02 x particles 1 mol 1 mol x particles Note: to convert between particles, mass, and volume, you have to go through moles.

Sec. 3: Percent Composition & Chemical Formulas  Percent Composition—The relative amount of each element in a compound The % of all elements in the compound must equal 100% The % of all elements in the compound must equal 100%  % mass of element = grams of element x 100 grams of compound grams of compound  Or  % mass of element = molar mass of element x 100 molar mass of compound

Example  An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the % composition?  First—add 8.20 g & 5.40 g to get the mass of the compound = g  % Mg = (mass of Mg/mass of compound) x 100 % Mg = 8.20g/13.60g x 100 = 60.3% % Mg = 8.20g/13.60g x 100 = 60.3%  % O = (mass of O / mass of compound) x 100 % O = 5.40/13.60 x 100 = 39.7% % O = 5.40/13.60 x 100 = 39.7%  Does this make sense? = = 100

Empirical Formulas  Empirical formula—gives the lowest whole- number ratio of the elements in a compound  An empirical formula may or may not be the same as a molecular formula. If the formulas are different, the molecular formula is a simple multiple of the empirical formula. If the formulas are different, the molecular formula is a simple multiple of the empirical formula.  Examples: The empirical formula for H 2 O 2 is HO The empirical formula for H 2 O 2 is HO For CO 2, the empirical & molecular formula are the same. For CO 2, the empirical & molecular formula are the same. C 6 H 6 and C 2 H 2 have the same empirical formula: CH C 6 H 6 and C 2 H 2 have the same empirical formula: CH

Molecular Formulas  You can determine the molecular formula if you know empirical formula and molar mass.  Divide the molar mass by the empirical formula mass.  Multiply this number by all subscripts in the empirical formula to get the molecular formula.

Example:  Find molecular formula with a molar mass of 60.0g and empirical formula of CH 4 N  1 st find the empirical formula mass 1 C, 4 H, 1 N 1 C, 4 H, 1 N 12 + (4 x 1) + 14 = 30 g 12 + (4 x 1) + 14 = 30 g  Then, divide molar mass by empirical mass 60.0g / 30 g = g / 30 g = 2  Multiply each element subscript by the this. C 2 H 8 N 2 C 2 H 8 N 2

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