Chapter 10 Answers

Mole-Particle Conversions #1) How many formula units are in 0.89 moles CaCl2? 1 mole = 6.02x1023 fu #2) A bottle of PbSO4 contains 4.25 x 1024 formula units of the compound. How many moles? 1 mole = 6.02x1023 fu

Mole-Particle Conversions #3) How many molecules are in 3.5 mole CH4? 1 mole = 6.02x1023 molecules #4) A tube with 3.68 moles of Neon contains how many atoms? 1 mole = 6.02x1023 atoms

Mole-Particle Conversions #5) 1 mole = 6.02x1023 molecules #6) 1 mole = 6.02x1023 ions #7) 1 mole = 6.02x1023 atoms

Mole-Mass Conversions #1) How many grams are in 0.75 moles of MgF2? 1 mole = molar mass (g) 1 mole = 62.3 g #2) A student measures 548.5 g Ca(NO3)2. How many moles? 1 mole = molar mass (g) 1 mole = 164.1 g

Mole-Mass Conversions #3) How many molecules are in 14.0 moles of H2O? 1 mole = 6.02x1023 molecules #4) A tube that contains 3.68 moles of Neon would be what mass? 1 mole = molar mass (g) 1 mole = 20.2 g

Mole-Mass Conversions #5) 1 mole = molar mass (g) 1 mole = 63.0 g #6) 1 mole = 6.02 x 1023 fu #7) 1 mole = molar mass (g) 1 mole = 63.5 g

1 step review 1 mole = molar mass (g) 1 mole = 181.0 g #1) You have 23 moles of Tantalum (Ta). How many grams is this? 1 mole = molar mass (g) 1 mole = 181.0 g #2) The label says that the bottle holds 500 grams of Al2(CO3)3. How many moles are there? 1 mole = molar mass (g) 1 mole = 234.0 g

1 step review 1 mole = 22.4 L 1 mole = molar mass (g) 1 mole = 107.9 g #3) How many moles of argon atoms are present in 11.2 L of argon gas at STP? 1 mole = 22.4 L #4) Your silver watchband masses out at 326 g. How many moles do you have? 1 mole = molar mass (g) 1 mole = 107.9 g

1 step review 1 mole = 6.02 x 1023 molecules 1 mole = 22.4 L #5) How many molecules are in 0.400 moles of N2O5? 1 mole = 6.02 x 1023 molecules #6) Determine the volume, in liters, occupied by 0.030 moles of a gas at STP. 1 mole = 22.4 L

1 step review #7) 1 mole = 342.0 g #8) 1 mole = 6.02x1023 atoms #9) 1 mole = 22.4 L

1 step review #10) 1 mole = 6.02x1023 fu #11) 1 mole = 60.0 g #12) 1 mole = 22.4 L

Two Step Conversions 1 mole = 6.02 x 1023 molecules 1 mole = 30.0 g #1) If you burn 6.10 x 1024 molecules of C2H6, what mass of ethane did you burn? 1 mole = 6.02 x 1023 molecules 1 mole = 30.0 g #2) A chemical reaction produces 1.38 L of HBr gas. How many grams? 1 mole = 22.4 L 1 mole = 80.9 g

Two Step Conversions 1 mole = 40.0g 1 mole = 6.02 x 1023 fu #3) How many formula units are in 13.5 g of NaOH? 1 mole = 40.0g 1 mole = 6.02 x 1023 fu #4) How many atoms are in 65 L of Argon gas at STP? 1 mole = 22.4 L 1 mole = 6.02 x 1023 atoms

Two Step Conversions 1 mole = 6.02 x 1023 molecules 1 mole = 22.4 L #5) How much volume would 8.4 x 1021 molecules of H2S gas occupy at STP? 1 mole = 6.02 x 1023 molecules 1 mole = 22.4 L #6) If I measure out 234 g of sugar (C6H22O11), how many molecules do I have? 1 mole = 270.0 g 1 mole = 6.02 x 1023 molecules

Two Step Conversions

All Mole Conversions 1 mole = 12.0g 1 mole = 6.02 x 1023 molecules #1) How many grams of carbon are in 3.5 moles? 1 mole = 12.0g #2) If I have 1.6 x 1024 molecules of FeCl3, how many grams do I have? 1 mole = 6.02 x 1023 molecules 1 mole = 162.3 g

All Mole Conversions 1 mole = 22.4 L 1 mole = 6.02 x 1023 molecules #3) How many grams are in a 5 L tank of NO2 (at STP)? 1 mole = 22.4 L 1 mole = 6.02 x 1023 molecules #4) How many moles of Zinc are in 1.65 x 1023 atoms? 1 mole = 6.02 x 1023 atoms

All Mole Conversions 1 mole = 22.4 L 1 mole = 4.0 g #5) There are about 700,000 L of Helium in a Macy’s Thanksgiving Day parade balloon. How many grams of Helium would this be (at STP)? 1 mole = 22.4 L 1 mole = 4.0 g #6) If I have 0.8 moles of sugar in my tea, how many molecules of sugar would I have? 1 mole = 6.02 x 1023 molecules

All Mole Conversions #7) 1 mole = 6.02 x 1023 atoms 1 mole = 22.4 L #8) 1 mole = 197.0 g 1 mole = 6.02 x 1023 atoms #9) 1 mole = 44.0 g 1 mole = 6.02 x 1023 molecules

All Mole Conversions

Percent Composition #1) NaOH Na – 23.0 x 1 = 23.0 23.0/40.0 x 100 = 57.5% Na O – 16.0 x 1 = 16.0 16.0/40.0 x 100 = 40.0% O H – 1.0 x 1 = 1.0 1.0/40.0 x 100 = 2.5% H 40.0g #2) (NH4)2SO4 N – 14.0 x 2 = 28.0 28.0/132.1 x 100 = 21.2% N H – 1.0 x 8 = 8.0 8.0/132.1 x 100 = 6.1% H S – 32. x 1 = 32.1 32.1/132.1 x 100 = 24.3% S O – 16.0 x 4 = 64.0 64.0/132.1 x 100 = 48.4% O 132.1g

Percent Composition Al – 1.30/2.45 x 100 = 53.1% Al #3) A sample of 2.45 g of aluminum oxide decomposes into 1.30 g of aluminum and 1.15 g of oxygen. What is the percent composition of the compound? Al – 1.30/2.45 x 100 = 53.1% Al O – 1.15/2.45 x 100 = 46.9% O #4) Find the percent composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen and 2.58 g of sulfur. whole = 1.94 + 0.48 + 2.58 = 5.00 C – 1.94/5.00 x 100 = 38.8% C H – 0.48/5.00 x 100 = 9.6% H S – 2.58/5.00 x 100 = 51.6% S

Percent Composition C – 12.0 x 6 = 72.0 72.0/180.0 x 100 = 40.0 % C #5) Determine the mass of Carbon in 120 g C6H12O6 C – 12.0 x 6 = 72.0 72.0/180.0 x 100 = 40.0 % C H – 1.0 x 12 = 12.0 O –16.0 x 6 = 96.0 180.0g 40.0 % of 120g 0.400 x 120 = 48g #6) What is the mass of magnesium in 23.2 g MgCl2? Mg – 24.3 x 1 = 24.3 24.3/95.3 x 100 = 25.5% Mg Cl – 35.5 x 2 = 71.0 95.3g 25.5 % of 23.2g 0.255 x 23.2 = 5.92g

Percent Comp. Back 1 mole = 148.3 g #1) Aluminum Chromate = Al2(CrO4)3 #2) How many moles are in 23 g of magnesium nitrate? = Mg(NO3)2 1 mole = 148.3 g

Percent Comp. Back 1 mole = 142.1 g 1 mole = 6.02x1023 molecules #3) How many grams are in 5.6 x 1022 molecules of sodium sulfate? Na2SO4 1 mole = 142.1 g #4) If you have 3.5 moles of CaBr2, how many molecules do you have? 1 mole = 6.02x1023 molecules

Percent Comp. Back #5) A balloon filled with 7.3 x 1025 atoms of helium at STP would take up how many liters? How many grams of helium would this be? #6) If your lungs could hold about 6 liters of oxygen (O2) at STP, how many grams of oxygen would that be?

Empirical Formulas #1) What is the empirical formula of a compound that contains 36.48 % Na; 25.44 % S; and 38.08 % O?

Empirical Formulas #2) #3)

Empirical Formulas #4) #5) 0.5000g – 0.2404 g = 0.2596g

Empirical Formulas Back

Empirical Formulas Back #3) C – 12.0 x 2 = 24.0 24.0/45.0 x 100 = 53.3 % C H – 1.0 x 5 = 5.0 5.0/45.0 x 100 = 11.1 % H O –16.0 x 1 = 16.0 16.0/45.0 x 100 = 35.6 % O 45.0g #4) S – 32.1 x 1 = 32.1 32.1/146.1 x 100 = 22.0 % S F –19.0 x 6 = 114.0 146.1g 22.0 % of 230g 0.220 x 230 = 50.6 g

Molecular Formulas #1) #2)

Molecular Formulas #3)

Ch 10 Problem Review #1) What is the atomic mass of Zn? 65.4 amu *Note units #2) What is the molar mass of Al2(SO4)3? 342.3 g *Note units #3) Find the mass of 0.159 moles of SiO2. #4) How many molecules are found in 3.65 moles of AgNO3?

Ch 10 Problem Review Zn – 65.4 x 1 = 65.4 65.4/183.4 x 100 = 35.7% Zn #5) If a container holds 6.5 L of CO2, what mass of gas is enclosed? #6) How many grams of mass of C2H6 would 2.37 x 1023 molecules of C2H6 contain? #7) If you have Zn(C2H3O2)2, what is the percent composition? Zn – 65.4 x 1 = 65.4 65.4/183.4 x 100 = 35.7% Zn C – 12.0 x 4 = 48.0 48.0/183.4 x 100 = 26.2% C H – 1.0. x 6 = 6.0 6.0/183.4 x 100 = 3.3% H O – 16.0 x 4 = 64.0 64.0/183.4 x 100 = 34.9% O 183.4g

Ch 10 Problem Review #8) Phenylalanine is an essential amino acid whose chemical composition is 65.5% C, 6.67% H, 8.48% N and 19.4% O. What is the empirical formula ?

Ch 10 Problem Review #9) Determine the empirical formula of a compound containing 24.74% potassium, 34.76% manganese, and 40.50% oxygen. #10) Find the empirical formula for a compound that contains 6.5 g potassium, 5.9 g chlorine and 8.0 g oxygen.

Ch 10 Problem Review 11.)

Chapter 10 Review What is a mole? -The number of atoms of an element equal to the number of atoms in exactly 12.0 g of carbon-12. (That number happens to be 6.02x1023) Why do we use the mole? -To simplify very large numbers Avagadro’s number? 6.02x1023 – It is the number of particles in a mole of any substance.

Chapter 10 Review Formula Mass? -The sum of the atomic masses in a compound (in amu’s) Molar Mass? -The mass of one mole of particles (in grams) Related? -Molar mass is equal to formula mass (or atomic mass) but in grams.

Chapter 10 Review Relationships -mass & moles: 1 mole = molar mass (g) -molecules/atoms/fu & moles: 1 mole = 6.02x1023 particles -volume & moles: 1 mole = 22.4 L Molar volume conditions? -STP (Standard Temp & Pressure) = 1 atm & 0°C

Chapter 10 Review Empirical Formula? -The simplest (reduced) formula (ratio of atoms) Molecular Formula? -Not reduced – so actual number of atoms present Related? -molecular formula = (empirical formula)x (molecular is empirical times some multiplier)

Chapter 10 Review Problems: 1.) 2.) 3.) 4.)

Chapter 10 Review Problems: 5.) Na – 23.0 x 2 = 46.0 46.0/134.0 x 100 = 34.3% Na C – 12.0 x 2 = 24.0 24.0/134.0 x 100 = 17.9% C O – 16.0 x 4 = 64.0 64.0/134.0 x 100 = 47.8% O 134.0g 6.) Copper(II)oxide = CuO Cu – 63.5 x 1 = 63.5 63.5/79.5 x 100 = 79.9% Cu O – 16.0 x 1 = 16.0 79.5g 0.799 x 325g = 259.7g

Chapter 10 Review 7.) 8.)C2HCl C – 12.0 x 2 = 24.0 H – 1.0 x 1 = 1.0 Cl – 35.5 x 1 = 35.5 60.5g

Ch 10 Review 9.)

Ch 10 Review 10.)

Ch 10 Practice Test Multiple Choice: B A C D Problems: 11. Na – 23.0 x 1 = 23.0 H – 1.0 x 1 = 1.0 C – 12.0 x 1 = 12.0 O – 16.0 x 3 = 48.0 84.0 amu Ba – 137.3 x 1 = 137.3 O – 16.0 x 2 = 32.0 H – 1.0 x 2 = 2.0 171.3 grams

Chapter 10 Review 13.) 14.) 15.) 16.)

Chapter 10 Review 17.) Fe – 2.232/3.192 x 100 = 69.9% Fe O – 0.96/3.192 x 100 = 30.1% O 18.)

Ch 10 Review 19.)

Chapter 10 Review 22.) 23.) 24.) H – 1.0 x 4 = 4.0 4.0/60.0 x 100 = 6.7% H C – 12.0 x 2 = 24.0 24.0/60.0 x 100 = 40.0% C O – 16.0 x 2 = 32.0 32.0/60.0 x 100 = 53.3% O 60.0g