Chapter 10 “Chemical Quantities” Yes, you will need a calculator for this chapter!

Section 10.1 Da Mole: A Measurement of Matter

How do we measure items?  Measure mass in.  Measure volume in.  Measure amount in.

What is the mole? We’re not talking about this kind of mole!

Moles (mol)  It is.  Defined as the number of carbon atoms in exactly 12 grams of carbon-12.  of representative particles.  6.02 x is called: number.  1 mole = 6.02 x = Avogadro’s number

Similar Words for amount Dozen:  1 dozen muffins = muffins  2 dozen puppies = puppies  36 m&m stuffed donuts = dozen donuts Mole :  1 mol muffins = muffins  2 mol puppies = puppies  2.4 x m&m stuffed donuts = mol donuts

What are Representative Particles?  The smallest pieces of a substance:  based on what we’re looking at 1)For a molecular compound: it is the. 2)For an ionic compound: it is the formula unit (ex: NaCl, MgS). 3)For an element: it is the. 1 mol of CO 2, 1 mol of NaCl, and 1 mol of H equal 6.02 x of that thing.

Measuring Moles  The mass on the PT is also the mass (g/mol) – mass (in grams) of 1 mole of that atom  =

Nitrogen Aluminum Zinc Find the Molar Mass (g/mol) of the following:

What about compound mass? In 1 mole of H 2 O, there are moles of H atoms and mole of O atoms  To find the mass of a compound: – Determine of each element present – the number times their mass (from the periodic table) – them up for the

Calculating Compound Mass Calculate the mass of magnesium carbonate, MgCO 3.

Practice Problem: What is the mass of one mole of CH 4 ? 1 mole of C = g/mol 4 mole of H x 1.01 g = 4.04g/mol 1 mole CH 4 =

10.1 Review 1.1 mol = particles mol MgCl 2 = ? particles MgCl 2 = x Fe particles = ? mol Fe =

4. Molar mass of BeF 2 = 5. Molar mass of C 6 H 12 O 6 = Be = 9.0 g/mol F = 19.0 g/mol C = 12.0 g/mol H = 1.0 g/mol O= 16.0 g/mol

Section 10.2 Mole-Mass and Mole-Volume Relationships

Since Molar Mass is…  Number of in 1 mole.  grams per mole (g/mol)  Use to make _ factors from these. - Use molar mass to convert to or of a substance.

For example How many moles is 5.69 g of NaOH? = How many grams in.53 mol of NaOH? =

Practice Problems:  How much would 2.34 moles of carbon weigh?  How many moles of magnesium is g of Mg?

The Mole-Volume Relationship  Under different circumstances, can.  Two things effect the volume of a gas: a) Temperature and b) **We need to compare all gases at the same temperature and pressure.

Standard Temperature and Pressure abbreviated (273K) and pressure At STP 1 mole of gas occupies 22.4 L= molar volume of any gas at STP

Practice Problems:  What is the volume of 4.59 mole of CO 2 gas at STP?  How many moles is 5.67 L of O 2 at STP?

What is the volume of 8.8 g of CH 4 gas at STP? =

Summary: These four items are all : a) 1 mole b) molar mass (in grams/mol) c) particles (atoms, molecules, or formula units) d) of a gas at **Thus, we can make factors from them.

Avog. # Practice problems:  How many molecules of CO 2 are in 4.6 moles of CO 2 ?  How many moles of water is in 5.87 x molecules?  How many atoms of carbon are in moles of Carbon?  How many moles is 7.78 x formula units of MgCl 2 ?

Mixed Practice Problems:  How many atoms of lithium is 1.0 g of Li?  How much would 3.45 x atoms of U weigh?  What is the volume of 10.0 g of CH 4 gas at STP?

Section 10.3 Percent Composition and Chemical Formulas

Percentage Composition Percentage by of element in a compound

Calculating Percent Composition of a Compound  Like all percent problems: part whole 1)Find mass of (from the periodic table) 2)Next, divide by total mass of ; then _ x 100 % = _

Percentage Composition Find the % composition of Cu 2 S. %Cu = %S =

%H 2 O = Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? Percentage Composition

Empirical Formula (EF) whole number of atoms in a compound C2H6C2H6C2H6C2H6 simplify subscripts

Empirical Formula (EF)  Just find whole number ratio C 6 H 12 O 6 CH 4 N  Formula is not just ratio of atoms, it is also ratio of.  In 1 mole of CO 2 there is mole of carbon and moles of oxygen.  In one molecule of CO 2 there is 1 atom of C and 2 atoms of O. = already lowest ratio.

Formulas (continued) Formulas for compounds are ALWAYS empirical (the lowest whole number ratio = be reduced). *Remember, we simplify beforehand Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

Formulas (continued) Formulas for molecular compounds be empirical (lowest whole number ratio). Molecular: C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O (Correct formula) (Lowest whole number ratio)

Formulas Empirical Formula (EF) = lowest whole number ratio of elements in a compound. Molecular Formula (MF) = the actual ratio of elements in a compound. The two can be the same. CH 2 = C 2 H 4 = C 3 H 6 =H 2 O =

Calculating EF We can get a ratio from the percent composition. 1.) Assume you have 100 g. -the percentage becomes grams (75.1% = 75.1 grams) 2.) Convert grams to moles. 3.) Find lowest whole number ratio by dividing by the smallest # of moles. **4.) If not a whole #, use a multiplier.

Example Calculate the empirical formula of a compound composed of % C, % H, and %N. Assume 100 g so: g C x 1mol C = mole C gC g H x 1mol H = mole H 1.01 gH g N x 1mol N = mole N gN Now divide each value by the smallest value

Example mol C = 1 mol C mol H = 5 mol H mol N = 1 mol N EF = C 1 H 5 N 1 = CH 5 N

Empirical Formula (EF) Find the empirical formula for a sample of 25.9% N and 74.1% O g 1 mol g = 1.85 mol N 74.1 g 1 mol g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5N2O5N2O5

EF Practice Find the empirical formula for a sample of 43.64% P and 56.36% O.

Molecular Formula (MF) “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6C2H6C2H6 empiricalformula molecularformula ?

Empirical to Molecular Since the empirical formula is the lowest ratio, the actual molecule would weigh more.  By a whole number multiple. **Divide the actual molar mass by the empirical formula mass.

Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol g/mol = 2 empirical mass = g/mol (CH 2 ) 2  C 2 H 4

Practice Problem: Caffeine (EF= C 4 H 5 N 2 O) has a molecular mass of 194 g. What is its molecular formula?