CHEMICAL QUANTITIES OR

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Presentation transcript:

CHEMICAL QUANTITIES OR Chapter 10 CHEMICAL QUANTITIES OR "Our Friend the Mole"

How you measure how much? You can measure mass (in grams), or volume (in liters), or you can count number of particles. We count number of particles in MOLES.

Moles Defined as the number of carbon atoms in exactly 12 grams of carbon-12. 1 mole is 6.02 x 1023 particles. Treat it like a very large dozen 6.02 x 1023 is called Avogadro's number.

Representative particles The smallest pieces of a substance. For an element, it usually is an atom. But for any diatomic element, it is a molecule For a molecular compound it is a molecule. For an ionic compound it is a formula unit.

Quick Review Molecular compounds are made of all nonmetallic elements. CO2 NO2 H2O Ionic compounds are made of ions. All ionic compounds begin with either a metal element or NH4 + 1 ion. NaCl, K2SO4 NH4 Br

What are the RP? Atom Formula unit Molecule Mg CaBr2 I2 CH4 Fe O2 NH4 NO3 NH3 Atom Formula unit Molecule

1mol of a substance= 6.02 x 1023 RP Conversion factor 1mol NH3 = 6.02 x 1023 M.C 1mol NaCl = 6.02 x 1023 F.Unit 1mol Mg = 6.02 x 1023 Atoms

Example 1: How many molecules of CO2 are the in 4.56 moles of CO2 ? Given: 4.56 moles CO2 Find: ? M.C (RP) Conversion factor: 1mol CO2= 6.02 x 10 23 M.C (RP) Given x conversion factor:

Example 2: How many moles of water is 5.87 x 1022 molecules of water? Given: 5.87 x 1022 molecules (RP) Find: ? moles water Conversion factor: 1mol H2 O= 6.02 x 10 23 M.C (RP) Given x conversion factor

Find: ? C atoms (atoms are particles but they are not RP for C6H12O6 ) Example 3: How many atoms of carbon are there in 1.23 moles of C6H12O6 ? Given: 1.23 moles C6H12O6 Find: ? C atoms (atoms are particles but they are not RP for C6H12O6 ) Conversion factors: Mole to M.C.(RP), then M.C (RP) to atoms. 1mol C6H12O6 = 6.02 x 10 23 M.C (RP) 1 C6H12O6 M.C = 6 C atoms

Given x conversion factor:

Molar Mass The mass of one any substance. Aka. Gram Formula mass

Gram Atomic Mass The mass of 1 mole of an element in grams.

Examples How much would 2.34 moles of carbon weigh?

Examples How many moles of magnesium in 4.61 g of Mg?

Examples How many atoms of lithium in 1.00 g of Li?

Examples How much would 3.45 x 1022 atoms of U weigh?

What about compounds? in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound determine the moles of the elements they have Find out how much they would weigh add them up

What about compounds? What is the mass of one mole of CH4? 1 mole of C = 12.01 g 4 mole of H x 1.01 g = 4.04g 1 mole CH4 = 12.01 + 4.04 = 16.05g

Molar Mass The mass of 1 mole What is the molar mass of Fe2O3? 2 moles of Fe x 55.85 g = 111.70 g 3 moles of O x 16.00 g = 48.00 g The GFM = 111.70 g + 48.00 g = 159.70g

Calculate the molar mass of the following C6H12O6 (NH4)3PO4

Finding moles of compounds Counting pieces by weighing Using Molar Mass Finding moles of compounds Counting pieces by weighing

Molar Mass The number of grams in 1 mole of atoms, formula units, or molecules. We can make conversion factors from these. To change grams of a compound to moles of a compound. Or moles to grams

For example need to change grams to moles for NaOH How many moles is 5.69 g of NaOH? need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

For example need to change grams to moles for NaOH How many moles is 5.69 g of NaOH? need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

Gases and the Mole

Gases Many of the chemicals we deal with are gases. They are difficult to weigh, so we’ll measure volume Need to know how many moles of gas we have. Two things affect the volume of a gas Temperature and pressure Compare at the same temp. and pressure.

Standard Temperature and Pressure Avogadro's Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles. 0ºC and 1 atmosphere pressure Abbreviated atm 273 K and 101.3 kPa kPa is kiloPascal

At Standard Temperature and Pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Used for conversion factors Moles to Liter and L to mol

What is the volume of 4.59 mole of CO2 gas at STP? Examples What is the volume of 4.59 mole of CO2 gas at STP?

Density of a gas at STP For a gas the units will be g / L To find the density we need the mass and the volume. Assume you have 1 mole, then the mass is the molar mass At STP the volume is 22.4 L. D = m/v = molar mass/molar volume

Examples Find the density of CO2 at STP. D = molar mass/ molar volume = 44.0g/ 22.4L=

Find the density of CH4 at STP. D = molar mass/ molar volume Example Find the density of CH4 at STP. D = molar mass/ molar volume = 16.0 g/ 22.4 L =

Calculating Molar Mass from Density Use the density of a gas at STP and the molar volume at STP (22.4L/mol) Molar mass = density (STP) x molar volume (STP)

Molar Mass WHAT IS THE MOLAR MASS OF A GAS WITH A DENSITY OF 1.964 G/L? MOLAR MASS = 1.964 G/L X 22.4 L /MOL = 44.0 G/MOL

Molar mass What is the molar mass of a gas with a density of 0.789 g/L? Molar mass = density x molar volume = 0.789 g/ L x 22.4 L/mol = g/ L

Your turn What is the density of NO2 gas at STP? What is the molar mass of an unknown gas with a density of 0.0785 g/L at STP?

Percent Composition Part x 100 % Whole Step 1: Find mass of each component and find total mass. Step 2: divide mass of each component by the total mass.

% composition from experimental data Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S. Mass of Ag: 29.0g Mass of S: 4.30g Total mass: 29.0 g + 4.30 g = 33.3g Ag% = Ag/total mass = 29.0 g/ 33.3g= S % = S/ total mass = 4.30 g/ 33.3g =

% composition from formula Calculate the percent composition of C2H4? Mass of C= 12.0 x 2 = 24.0 g Mass of the whole compound= 24.0 g + 4x 1.01 = 28.0g % C= 24.0 g/ 28.0 g x 100% = % H = 4 H / C2H4 x 100% =

Calculate mass of each component from % composition Mass of element A = % of element A x total mass

Example How many grams of aluminum is present in 450 g of aluminum carbonate Al2 (CO3)3 ? Step 1: find % of Al from formula Mass of aluminum in the formula: 2 x 27g Total formula mass: 2 x 27 + (12 + 3 x 16) x 3 = g Al % = 2 x 27 / Step 2: Mass of Al = x 450 g =

How many grams of iron are there in 3.45 g of FeS? Step 1: find % of Fe from the formula Mass of Fe in the formula: 56 g Total formula mass of FeS = 56+32 =88 g Fe% = 56 /88 x 100% = Step 2: Mass of iron = % of iron x total mass

The Empirical and Molecular Formula EF: The lowest whole number ratio of elements in a compound. MF: the actual ratio of elements in a compound. The two sometimes are the same. MF: C 2H4 EF: CH2 MF: C3H6 EF: CH2 MF: H 2O EF: H2O

Interpreting EF CH2 1 C atom combines with 2 H atom 1 dozen C atoms, 2 dozens H atoms 1 mole C atoms, 2 moles H atoms 2 moles N atoms combines with 5 moles of H atoms, what is EF? N2 H5

Finding EF 4 moles of H atoms combines with 2 moles of O atoms, what is EF? H4O2 change to H2 O Finding EF is to find the lowest whole number mole ratio of elements in the compound

Calculating Empirical Formulas from % composition Step 1: Assume you have 100 g of the compound. The percentages become grams. Step 2: Turn grams to moles using gram atomic mass of each element. Step 3: Divide all the answers by the smallest moles.

. 33 or.34 mole, multiply all answers by 3 Note: after step 3, if your answers are still not whole numbers, take the following steps to change your answers to whole numbers: . 33 or.34 mole, multiply all answers by 3 .25 mole, multiply all answers by 4 . 5 mole, multiply all answers by 2

Calculate the empirical formula of a compound composed of 38 Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Step 1: Assume 100g, so we have 38,67g C, 16.22g H, 45.11g N Step 2: 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.1 mole H 1.01g H 45.11 g N x 1mol N = 3.220 mole N 14.01 gN

Step 3: 3.220 mol C = 1 mol C 3.220 mol EF is C1H5N1 16.1 mol H = 5 mol H 3.220 mol 3.220 mol N = 1 mol N EF is C1H5N1

Calculating EF A compound is analyzed and found to contain 25.9% nitrogen and 74.1 % oxygen. What is the empirical formula of the compound? Step 1: 25.9 g N 74.1g O Step 2: 25.9 g N x1mol/ 14 g = 1.85 mol N Step 3: 74.1 g O x 1mol/16g = 4.63 mol O Step 4: 1.85 mol N /1.85 = 1 mol N 4.63 mol O/ 1.85 = 2.50 mol O Step 5: 1mol N x 2 = 2 mol N 2.5 mol O x 2 = 5 mol O N2 O 5

Practice Problems P310: Q 36, 37 P312: Q 45, 46

Empirical to molecular Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

Calculating MF A compound has an empirical formula of ClCH2 and a molar mass of 98.96 g/mol. What is its molecular formula? A compound has an empirical formula of CH2O and a molar mass of 180.0 g/mol. What is its molecular formula?

Percent to molecular Take the percent x the molar mass This gives you mass in one mole of the compound Change this to moles You will get whole numbers These are the subscripts Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. It has a molar mass of 194 g. What is its molecular formula?

Example Ibuprofen is 75.69 % C, 8.80 % H, 15.51 % O, and has a molar mass of about 207 g/mol. What is its molecular formula?