 #  Objectives ◦ Describe how to convert the mass of a substance to the number of moles of a substance, and vice versa. ◦ Identify the volume of a quantity.

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 Objectives ◦ Describe how to convert the mass of a substance to the number of moles of a substance, and vice versa. ◦ Identify the volume of a quantity of gas at STP.

 A mole is a unit of measurement ◦ Kind of like a dozen = 12, or a ream = 500  A mole is a set number of particles, molecules, atoms, or just whatever it is we are talking about. ◦ That number is 6.02 x 10 23

 If we wish to find the moles or particles of something we follow two easy set ups. ◦ First if we are looking for moles:  (Number of Particles Given) x ( 1 mole/ 6.02 x 10 23 particles)= The number of Moles ◦ Second, if we are looking for the number of particles:  (Number of Moles Given) x (6.02 x 10 23 particles/1 Moles)= The number of Particles

 If we wish to convert between the moles and mass of a compound, we need the molar mass. ◦ For Example:  If we wanted to see how many grams are present in 3 moles of NaCl, we simply do the following  (Number of Moles) x (Molar Mass/ 1 mole)  3 moles of NaCl x (58.5 g/mole of NaCl) = 176 grams of NaCl

 To find moles from mass: ◦ (Mass of Compound Given) x (1 mole/Molar Mass) = Number of Moles

 Find the mass of :  4.5 x 10 -3 moles of C 20 H 42  2.5 moles of Fe(OH) 2  Calculate the Number of moles of:  3.7 x 10 -1 grams of Boron  75.0 grams of N 2 O 3

 Avogadro’s Hypothesis states that equal volumes of gases at the same temperature and pressure contain equal number of particles.  When we try and calculate using a gas, we need to confirm one item. That is STP, or standard temperature and pressure. At these perfect conditions we can progress in the calculations

 As we learned before, 1 mole of anything is equal to 6.02 x 10 23  For gases, 1 mole is also equal to 22.4 L ◦ In other words a mole of ANY gas will occupy 22.4 L

 If we have the following:  0.375 moles of O 2 gas, what volume will it occupy  Volume of Gas = (Moles of Gas) x (22.4 L/1 mole)  With this we get the following  Volume of O 2 = (0.375 mole) x (22.4 L /1 mole)  Volume of O 2 = 8.40 L

 Calculate Volume of the given gases  3.2 x 10 -3 mol of CO 2  3.70 mol N 2  1.25 mol He  0.335 mol C 2 H 6

 Describe how to calculate the percent by mass of an element in a compound  Interpret an empirical formula  Distinguish between empirical and molecular formulas.

 How many people are in the room right now?  How many are Male? Female?  What percentage of the whole is Male? Female?

 The percent by mass of an element in a compound is the number of grams of the compound multiplied by 100%  Percent Composition is the percent by mass of each element in a compound.

 K 2 CrO 7  How many Potassiums, Chromiums, and Oxygen do we have?  Whats the total molar mass of the compound?  What is the percent composition of the elements compared to the whole?

 When a 13.60 gram sample of a compound containing magnesium and oxygen is decomposed,5.40 grams of oxygen is obtained. What is the percent composition of this compound?

 We use the subscripts to tell us the amount of each element in a mole of compound, which in turn can tell us the mass, and percent composition.  Similar to when we solved for K 2 CrO 7, let us try and find the percent composition for propane, C 3 H 6.

 When we looked at propane we could see a few things: ◦ The molar mass is 44.0 grams ◦ The mass of carbon was 36 grams ◦ The mass of hydrogen was 8 grams ◦ The Percent composition of carbon is 81.8% ◦ The Percent composition of hydrogen is 18.2%  But what would the masses be if we had 82 grams of propane?

 An empirical formula is the lowest whole number ratio of the atoms in a compound. ◦ CH ◦ CH 2 O  A molecular formula is either the same as the empirical formula, or it appears as a multiple from the empirical formula.  C 2 H 2  CH 2 O

 You are told the empirical formula of a substance is CH 4 N, and that you have 60 g/mol of substance.  What is the molecular formula?

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