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CHEMICAL QUANTITIES THE MOLE CONCEPT AND ITS APPLICATIONS

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C HEMICAL Q UANTITIES O BJECTIVES 1. S TUDENTS WILL BE ABLE TO U NDERSTAND AND KNOW THE DEFINITIONS OF THE MOLE AND A VOGADRO ’ S N UMBER RELATIONSHIP BETWEEN MOLES AND MASS RELATIONSHIP BETWEEN MOLES AND PARTICLES C ALCULATE THE FORMULA MASS OF A COMPOUND GIVEN THE PERIODIC TABLE C ALCULATE THE MASS PERCENT COMPOSITION OF A COMPOUND GIVEN THE FORMULA, FORMULA MASS, AND PERIODIC TABLE.

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D RILL & O BJECTIVES 1. How many eggs are there in a dozen? 1. If I bought 3 dozen eggs, how many eggs do I have? 2. What are some ways that you can measure things? (such as the amount of a substance) O BJECTIVES 1. S TUDENTS WILL BE ABLE TO U NDERSTAND AND KNOW THE DEFINITIONS OF THE MOLE AND A VOGADRO ’ S N UMBER RELATIONSHIP BETWEEN MOLES AND MASS RELATIONSHIP BETWEEN MOLES AND PARTICLES C ALCULATE THE FORMULA MASS OF A COMPOUND GIVEN THE PERIODIC TABLE C ALCULATE THE MASS PERCENT COMPOSITION OF A COMPOUND GIVEN THE FORMULA, FORMULA MASS, AND PERIODIC TABLE.

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C HEMICAL Q UANTITIES How many eggs are there in a dozen? 12!

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C HEMICAL Q UANTITIES How many roses are there in a dozen roses? 12!

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C HEMICAL Q UANTITIES Does it matter what the _____________ is?

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C HEMICAL Q UANTITIES Does it matter what the _____________ is? NO! A Dozen is a dozen! No matter if it’s flowers, eggs, bagels, ect. It represents a number of _______

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C HEMICAL Q UANTITIES In Chemistry the same concept is valid, Let’s talk about “The Mole” and how it relates to Chemistry…

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T HE M OLE ?

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The Mole?

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T HE M OLE ? “The Mole” or “A Mole” is an amount of a substance There are 6.02 x10 23 atoms (representative particles of a substance) in 1 Mole (of that substance). This number (6.02 x10 23 ) is called Avogadro’s number Named after Amedeo Avogadro di Quarenga (1776 – 1856), an Italian scientist.

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T HE M OLE ! Let’s put this into perspective How much is 1 mol of a substance? 6.023 x 10 23 !!

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T HE M OLE ! IF you had 6.02 X 10 23 Watermelon Seeds…

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T HE M OLE ! IF you had 6.02 X 10 23 Watermelon Seeds… it would be found inside a watermelon slightly larger than the moon!

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T HE M OLE ! If you had 6.02 X 10 23 Grains of Sand…

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T HE M OLE ! IF you had 6.02 X 10 23 Grains of Sand… it would be more than ALL the sand on Miami Beach.

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T HE M OLE ? A mole is the amount of substance or representative particles of any substance. Therefore, as we have just seen 1 mol of N 2 there are 6.02x10 23 MOLECULES of N 2 1 mol of C 12 H 22 O 11 has how many molecules? 6.02x10 23 can be used for “atoms” if they are the same element OR It can be used for “molecules” for many elements that are together, via compounds.

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C HEMICAL Q UANTITIES How many moles of Magnesium is 3.01x10 22 atoms of magnesium? Step 1 - What do we know? We know that there are 3.01x10 22 atoms of Mg Step 2 - What do we want know? We want to know the number of moles of Mg Step 3 – What do we know that can be used for these conversions? We know that there are 6.02x10 23 atoms in 1 mol Step 4 – Solve the problem

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C HEMICAL Q UANTITIES How many moles of Magnesium is 3.01x10 22 atoms of magnesium? 3.01x10 22 atoms Mg X 1 mol Mg 6.02x10 23 atoms of Mg = 5.00x10 -2 mol Mg

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C HEMICAL Q UANTITIES Individual Problems How many moles are 1.20x10 25 atoms of phosphorus? How many atoms are in 0.750 mol of Zinc? How many molecules are there in 4 mol of glucose, C 6 H 12 O 6 ? How many molecules are there in 0.44 mol N 2 O 5

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C HEMICAL Q UANTITIES Now there is a difference between asking how many atoms are there in the entire compound and how many atoms of one of the compound there are. We said that there are the same number or atoms in a compound or a diatomic molecule. However, there is a difference between these questions… How many atoms are there in aluminum fluoride? VS. How many fluoride ions are there in aluminum fluoride?

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C HEMICAL Q UANTITIES So let’s answer the question. How many fluoride ions are there in 1.46 mol of aluminum fluoride? If we follow the four steps that we went through prior

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C HEMICAL Q UANTITIES How many fluoride ions are there in 1.46 mol of aluminum fluoride? Step 1 - What do we know? We know that there are 1.46 mol of aluminum fluoride Step 2 - What do we want know? We want to know the number of F - Step 3 – What do we know that can be used for these conversions? We know that there are 6.02x10 23 atoms in 1 mol BUT is that enough information?? NO!

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C HEMICAL Q UANTITIES How many fluoride ions are there in 1.46 mol of aluminum fluoride? We want to know the number of F - But how can we find out the number of fluoride ions we have from what is given? We can get the information we need by writing out the chemical formula of the compound. AlF 3

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C HEMICAL Q UANTITIES 1.46 mol AlF 3 X 1 mol AlF 3 6.02x10 23 Formula units of AlF 3 How many fluoride ions are there in 1.46 mol of aluminum fluoride? X 1 formula unit AlF 3 3 F - ions = 26.3676 x 10 23 F - ions

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I NDIVIDUAL W ORK How many ammonium ions are in 0.036 mol ammonium phosphate, (NH 4 ) 3 PO 4 ?

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DRILL What is Avogadro's number? How many particles are there in 1 mol of a substance? How many atoms are there in 1.45 mol of Na?

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C HEMICAL Q UANTITIES T HE G RAM F ORMULA M ASS

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C HEMICAL Q UANTITIES Chemists have defined the gram atomic mass as the number of grams of an element that is numerically equal to the atomic mass in amu The Gram atomic mass is the mass of one mole of atoms of a monatomic element This can also be described as MOLAR MASS – in place of gram formula mass to refer to the mass of a mole of any element or compound. I will be using the term Molar Mass more frequently

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C HEMICAL Q UANTITIES Atomic Mass Number

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C HEMICAL Q UANTITIES This number tells you how many grams of this element there are in 1 mol of the element Therefore, there are 47.88 grams/mol of Ti Or there are 47.88 grams of Ti IN 1 mol of Ti

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C HEMICAL Q UANTITIES BCNOF 10.811g12.0107g14.0067g15.9994g18.998g The molar mass in grams/mol

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C HEMICAL Q UANTITIES You can calculate the total molecular weight of a molecule by adding up the molar masses of each element. The molar mass of the molecule SO 4 is equal to S = 32.065 g/mol x 1 S = 32.065 g/mol O = 15.9994 g/mol x 4 O = 63.9976 g/mol SO 4 = 96.0626 g/mol I will specify the amount of significant figures needed in your answer

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I NDIVIDUAL W ORK What is the molar mass of the following compounds? REMEMBER YOUR UNITS! 1. PCl 3 2. Sodium Carbonate 3. C 8 H 18 4. Aluminum Sulfate 5. (NH 4 ) 2 CO 3

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C HEMICAL Q UANTITIES Why is this important? This is important because it will allow us to covert from moles to grams and grams to moles, allowing for quantitative experimentation. Let’s do a calculation… How many grams are in 7.20 mol of dinitrogen trioxide?

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C HEMICAL Q UANTITIES How many grams are in 7.20 mol of dinitrogen trioxide? Step 1 – add up the total molar mass of the compound If it is in word form, write the chemical formula Step 2 – Set up the proper conversion factors Step 3 – Solve

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C HEMICAL Q UANTITIES How many grams are in 7.20 mol of dinitrogen trioxide? Step 1 – add up the total molar mass of the compound If it is in word form, write the chemical formula N 2 O 3 N = 14.0 g/mol x 2 N = 28.0 g/mol O = 16.0 g/mol x 3 O = 48.0 g/mol N 2 O 3 = 76.0 g/mol

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C HEMICAL Q UANTITIES How many grams are in 7.20 mol of dinitrogen trioxide? Step 2 – Set up the proper conversion factors Step 3 – Solve

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I NDIVIDUAL W ORK Find the mass of the following 1. 3.32 mol K 2. 5.08 mol Ca(NO 3 ) 2 3. 4.52x10 -3 mol K 2 CO 3 Find the number moles of the following 1. 0.000264g Li 2 HPO 4 2. 847g (NH 4 ) 2 CO 3 3. 195g calcium nitrate

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E XIT T ICKET Does 54.938 g of Mn have the same number of moles as that of 112.411 g of Cd? Explain why or why not.

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DRILL What is Avogadro’s number? How many moles are there in 245 kg of CH 2 COOH? Does 54.938 g of Mn have the same number of moles as that of 112.411 g of Cd? Explain why or why not.

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Group Work Find the mass (g) of the following: 1. 10.0 mol Cr 2. 2.20x10 -3 mol Sn 3. 0.720 mol Be 4. 2.40 mol N 2 5. 4.52x10 -3 mol C 20 H 42 6. 0.0112 mol Potassium Carbonate 1. Find the number of moles of the following: 1. 72.0 g Ar 2. 3.70x10 -1 g B 3. 333 g Tin (II) Fluoride 4. 7.21x10 -2 g He 5. 27.4 g TiO 2

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C HEMICAL Q UANTITIES T HE V OLUME OF A M OLE OF G AS

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C HEMICAL Q UANTITIES How would we measure the amount of moles in a gas?

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C HEMICAL Q UANTITIES The volume of a gas is usually measured at STP (Standard Temperature and Pressure) STP conditions Temperature 0°C Pressure 1 atmosphere (atm)

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C HEMICAL Q UANTITIES At STP conditions 1 mol of ANY gas occupies a volume of 22.4 Liters (L) 22.4 Liters of a gas / 1 mol of the gas 22.4 L of a gas contains 6.02 x 10 23 representative particles of that gas This is known as the molar volume of a gas

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C HEMICAL Q UANTITIES What were the units of the molar volume of a gas? Liters What kind of unit is liters? VOLUME Therefore, 1 mol of any gas occupies the same volume not mass

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C HEMICAL Q UANTITIES He Ne CO 2 22.4 L He 6.02 x 10 23 molecules of He 4g 22.4 L N 2 6.02 x 10 23 molecules of N 2 28g of N 2 22.4 L CO 2 6.02 x 10 23 molecules of CO 2 44g of CO 2

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I NDIVIDUAL W ORK What is the volume at STP of these gases? 5.40 mol O 2 3.20 x 10 -2 mol CO 2 Assuming STP conditions, how many moles are there in these volumes 74.6 L SO 2 5.78x10 -2 N 2

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C HEMICAL Q UANTITIES Because the molar volume of a gas is a VOLUME, we can relate density (mass over volume) of a gas to determine the mass of the gas. If we have a gas that has a density of 1.964 g/L we can multiply the density with the molar volume of a gas (22.4 L) to calculate the mass of the gas. 1.964 x 22.4 L = 44.0g

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G ROUP W ORK The densities of gases A, B, & C are 1.25 g/L, 2.86 g/L, and 0.714 g/L, respectively. Calculate the molar mass of each of these substances and compare them to the molar mass of ammonia, sulfur dioxide, chlorine, nitrogen, and methane Identify the gasses A, B, & C

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C HEMICAL Q UANTITIES Atoms (6.02 x 10 23 )1mol Molar Mass Solids and Liquids 1 mol22.4 LGases

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H OMEWORK PROBLEMS C OPY THEM DOWN ! 1. Calculate the volume of each of these gases at STP 1. 9.6 mol He 2. 4.8 mol N 2 2. How many moles is each of the following, assuming STP 1. 56.o L N 2 O 2. 0.224 L O 2 3. Find each of the following quantities 1. The mass of 18.0L of CH 4 (STP) 2. The volume in liter, of 835 g of SO 3 (STP)

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D RILL What are STANDARD conditions? How many liters are there in 1 mol of gas? Does it matter which gas it is? Why or why not?

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C HEMICAL Q UANTITIES P ERCENT C OMPOSITION

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C HEMICAL Q UANTITIES - % C OMPOSITION Calculating Percent Composition Percent Composition The percent by mass of each element in a compound They must add up to 100% Ex. K 2 CrO 4 40.3% K 26.8% Cr 32.9% O 100% Total

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C HEMICAL Q UANTITIES - % C OMPOSITION Percent Mass of an element in a compound is the number of grams of the element divided by the grams of the compound, multiplied by 100% % mass =x 100% Grams of element Grams of compound

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C HEMICAL Q UANTITIES - % C OMPOSITION Let’s look at our example again K 2 CrO 4 STEP 1 Calculate the Total Molecular Mass of K 2 CrO 4 – Go ahead a calculate that now K = 39 g x 2K = 78 g Cr = 52 g O = 16 g x 4O = 64g K 2 CrO 4 = 194g

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C HEMICAL Q UANTITIES - % C OMPOSITION Let’s look at our example again K 2 CrO 4 STEP 2 Using the individual molar mass of each element, divide each element with the total molar mass Go ahead and calculate that now with 3 Significant figures K = 39 g x 2K = 78 g K ÷194g =.402 Cr = 52 g ÷194g = 0.268 O = 16 g x 4O = 64g ÷194g = 0.330

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C HEMICAL Q UANTITIES - % C OMPOSITION Let’s look at our example again K 2 CrO 4 STEP 3 Multiply the number calculated from Step 2 and multiply by 100% Go ahead and calculate that now K = 39 g x 2K = 78 g K ÷194g =.402 x 100% = 40.2% K Cr = 52 g ÷194g = 0.268 x 100% = 26.8% Cr O = 16 g x 4O = 64g ÷194g = 0.330 x 100% = 33% O

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C HEMICAL Q UANTITIES - % C OMPOSITION I NDIVIDUAL W ORK 1. Calculate the mass of carbon in 82g in C 3 H 8 2. Calculate the percent composition of each of these compounds. 1. Propane, C 3 H 8 2. Sodium bisulfate, NaHSO 4 3. Calcium acetate, Ca(C 2 H 3 O 2 ) 2 4. Hydrogen cyanide, HCN

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C LOSURE Take the remainder of this period and write down in complete sentences what you learned today, approximately 2-3 paragraphs.

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DRILL What is percent composition? What is the formula for percent composition?

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DRILL Calculate the mass of oxygen in 142g of Sodium bisulfate, NaHSO 4

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C HEMICAL Q UANTITIES E MPIRICAL F ORMULAS

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS Empirical Formula Gives the lowest whole number ratio of the elements in a compound. This ratio is not necessarily the molecular formula! Calculating the empirical formula is going to require the use of almost everything you’ve learned so far. Each element in the compound would require different amounts depending on the mass and % composition

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS Steps to solve for an Empirical Formula 1. Write down the base formula A x B Y 2. Convert whatever you have into moles using dimensional analysis 3. Divide each element by the lowest number of moles in the compound 4. Round if necessary to the nearest 10 ths place and/or multiply by a number to get a whole number ratio 5. Write down the Empirical Formula

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 1 - Write down the base formula A x B Y … N x O y

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 2 - Convert whatever you have into moles using dimensional analysis Because this is a % composition, we know that everything has to add up to 100%; therefore, we can assume that if we had 100g total of the compound, there would be 25.9g of N and 74.1g of O

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 2 - Convert whatever you have into moles using dimensional analysis 1 mol N 14.0 g N 1 mol O 16.0 g O 25.9g of N x = 1.85 mol N 74.1g of O x= 4.63 mol O

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 3 - Divide each element by the lowest number of moles in the compound 25.9g of N x= 1.85 mol N 74.1g of O x= 4.63 mol O 1 mol N 14.0 g N 1 mol O 16.0 g O ÷ 1.85 mol = 1 mol N ÷ 1.85 mol = 2.5 mol O

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 4 - Round if necessary to the nearest 10 ths place and/or multiply by a number to get a whole number ratio 1 mol N x ____ = 2.5 mol O x ____ = 2 2 2 mol N 5 mol O

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C HEMICAL Q UANTITIES - E MPIRICAL F ORMULAS What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 5 - Write down the Empirical Formula 2 mol N5 mol O N 2 O 5

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C HEMICAL Q UANTITIES - M OLECULAR F ORMULAS The Molecular formula is the actual formula of a molecular compound While the empirical formula give us the base ratio of the compound, the molecular formula will give the actual molecular formula by using the molar mass. x (empirical formula) = molecular formula Ratio of the compounds

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Chemical Quantities- Molecular Formulas Therefore, if you have the molar mass of the compound and divide it by the molar mass of the empirical formula, you will get the ratio between the two. Multiply the ratio to each element of the compound to get the the molecular formula.

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Chemical Quantities- Molecular Formulas Ex. If the empirical formula was calculated to be CH 4 N and the known molecular formula molar mass is = to 60g 1 st – Take the molar mass of the empirical formula CH 4 N In this case it is calculated to be 30g/mol 2 nd – Divide the known molar mass of 60g/mol and divide it by 30 g/mol 60/30 = 2 3 rd – Take the ratio of the two molar masses and then multiply it by each element of the compound C 1x2 H 4x2 N 1x2 = C 2 H 8 N 2

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Chemical Quantities- Molecular Formulas Write the Molecular Formula for the problem below The compound methyl butanoate has a percent composition of 58.8% C, 9.8% H, 31.4%O. If its molecular mass is 102 g/mol, what is it’s molecular formula? Write the Empirical Formula for the following 1. 79.8% C20.2% H 2. 67.6% Hg 10.8% S 21.6% O 3. 27.59% C1.15% H16.09%N55.17%O 4. 17.6% Na39.7%Cr42.7%O

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DRILL What is the difference between an empirical formula and molecular formula?

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Chemical Quantities – Review Determine the Empirical Formula with the parameters below 1. 71.72% Cl, 16.16% O, and 12.12%C What is the molecular formula for the compound below? The compound’s empirical formula and molar mass is given Below 2. HgCl, 472.2 g/mol Determine the molecular formula for the compound 3. 94.1%O and 5.9% H ; molar mass = 34g Find the empirical formula for each compound from its percent composition 4. 72.4% Fe and 27.6% O 5. 52.8% Sn, 12.4% Fe, 16.0% C, and 18.8% N

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