NOTES – MOLAR MASS AND MOLE CONVERSIONS



Name: ________________________________________Kaspriskie/GraubChapter 10: MOLAR MASS AND PERCENT COMPOSTION5181600-508000MOLAR MASS OF COMPOUNDSWhat is the molar mass of a substance?The molar mass of a substance is the mass (in grams) of 1 mole of that substance (or, the mass of 6.02 X 1023 particles of that substance) REVIEW: What is the molar mass of the element Carbon?12.01 g/molWhat is the molar mass of Nitrogen gas (N2)?28.02 g/mol59944090805To calculate MOLAR MASS of a compound…..If necessary, write the formula of the compound from the nameFind atomic mass of each elementMULTIPLYeach atomic mass by the appropriate subscriptsADD 00To calculate MOLAR MASS of a compound…..If necessary, write the formula of the compound from the nameFind atomic mass of each elementMULTIPLYeach atomic mass by the appropriate subscriptsADD EXAMPLES:Find molar mass of water. FORMULA H2O2 mol H 2(1.01g/mol) = 2.02g/mol1 mol O 1(16.00g/mol)= 16.00g/mol18.02g/mol So, the Molar Mass of H2O = 18.02 g/mol Find molar mass of sulfuric acidFORMULA H2SO42 mol H 2(1.01 g/mol) = 2.02 g/mol 1 mol S 1(32.07 g/mol)= 32.07 g/mol 4 mol O 4(16.00 g/mol)= 64.00 g/mol98.09 g/molSo, the Molar Mass of H2SO4 = 98.09 g/mol YOU TRY IT sodium hydroxide FORMULA: ____NaOH_____Na1+ (OH)1- Na: 1 x 23.0 g/mol= 23.0 g/mol O: 1 x 16.0 g/ mol= 16.0 g/ mol H: 1 x 1.0 g/mol = 1.0 g/mol 40.0 g/mol 44100759906000magnesium hydroxide FORMULA: __Mg(OH)2_____ OH OHMg2+ (OH)1-Mg: 1 x 24.3 g/mol= 24.3 g/mol O: 2 x 16.0 g/mol= 32.0 g/mol H: 2 x 1.0 g/mol = 2.0 g/mol 58.3 g/molcalcium cyanideFORMULA: ___Ca(CN)2_____Ca2+ (CN)1-Ca: 1 x 40.1 g/mol= 40.1 g/mol C: 1 x 12.0 g/mol= 24.0 g/mol N: 1 x 14.0 g/mol= 28.0 g/mol 92.1 g/mol45624759398000magnesium phosphateFORMULA: ___Mg3(PO4)2___ PO4 PO4Mg2+ (PO4)3-Mg: 3 x 24.3 g/mol= 72.9 g/mol P: 2 x 31.0 g/mol= 62.0 g/mol O: 8 x 16.0 g/mol= 128.0 g/mol 262.9 g/mol iron(III) dichromate FORMULA: Fe2(Cr2O7)3____Fe3+ (Cr2O7)2-Fe: 2 x 55.8 g/mol = 111.6 g/molCr: 6 x 52.0 g/mol = 312.0 g/mol O: 21 x 16.0 g/mol= 336.0 g/mol 759.6 g/mol ammonium hydroxideFORMULA: __NH4OH____N: 1 x 14.0 g/mol = 14.0 g/mol(NH4)1+ (OH)1-H: 5 x 1.0 g/mol = 5.0 g/molO: 1 x 16.0 g/mol = 16.0 g/mol 35.0 g/moldinitrogen pentoxide FORMULA: _N2O5_____N: 2 x 14.0 g/mol = 28.0 g/molO: 5 x 16.0 g/mol = 80.0 g/mol 108.0 g/molPER CENT COMPOSITION OF AN ELEMENT IN A COMPOUND4196715109855-78105243840To calculate PERCENT COMPOSITION…….Write the formula of the compoundFind the molar mass of the each elementFind the TOTAL MOLAR MASS of the compoundDIVIDE the molar mass of each element by TOTAL MOLAR MASS of the compound.MULTIPLY each by 100 to find percent of each00To calculate PERCENT COMPOSITION…….Write the formula of the compoundFind the molar mass of the each elementFind the TOTAL MOLAR MASS of the compoundDIVIDE the molar mass of each element by TOTAL MOLAR MASS of the compound.MULTIPLY each by 100 to find percent of eachEXAMPLES: Find the per cent composition of calcium cyanide: FORMULA: ____Ca(CN)2_________Ca: 1 x 40.08= 40.08 g/mol% Ca = _43.52%____C : 2 x 12.01 g/mol= 24.02 g/mol N: 2 x 14.01 g/mol= 28.02 g/mol% C = _26.07%___ 92.12 g/mol % N = _30.42%____ *****From per cent composition, we can determine the mass of an element in a given sample of that compound: How many grams of calcium are in 40.0g of calcium cyanide?40.0 g Ca(CN)2 x 40.08 g/mol Ca = 17.40 g Ca 92.12 g/mol Ca(CN)2 Find the per cent composition of magnesium phosphate:FORMULA: __Mg3(PO4)2___________ Mg: 3 x 24.31 g/mol =72.93 g/mol% Mg = 27.74%___ P: 2 x 30.97 g/mol =61.94 g/mol% P = _23.56%____O: 8 x 16.00 g/mol = 128.00 g/mol% O = _48.69 %____ 262.87 g/molHow many grams of magnesium are in 350g of magnesium phosphate? 350 g Mg3(PO4)2 x 72.93- g/mol Mg = 97.09 g Mg 262.87 g/mol Mg3(PO4)2*****Extension: From percent composition, we can determine the mass of a given sample of that compound if we know the mass of an element in that sample: What mass of magnesium phosphate contains 15g of magnesium?15 g Mg x 262.87 g/mol Mg3(PO4)2 = 54.07 g Mg3(PO4)2 72.93g/mol MgYOU TRY SOME: 1. Find the molar mass of each compound (review):___73.89 g/mol______29622751123950lithium carbonateLi: 2 x 6.94 g/molLi 1+ (CO3)2-C: 1 x 12.01 g/molLi2 CO3 O: 3 x 16.00 g/mol31051501130300 calcium nitrateCa: 1 x 40.08 g/mol ___164.1 g/mol_______Ca2+ (NO3)1-N: 2 x 14.01 g/molCa(NO3)2O: 6 x 16.00 g/mol31623001225550 tin (IV) sulfateSn: 1 x 118.71 g/mol____310.85 g/mol_________Sn4+ (SO4)2-S: 2 x 32.07 g/molSn(SO4)2O: 8 x 16.00 g/mol2. Find the percent composition of tin (IV) sulfate.%Sn: 118.71 g/mol x 100 = 38.19% 310.85 g/mol% Sn = _38.19% ___%S: 64.14 g/mol x 100 = 20.63% 310.85 g/mol % S = _20.63% ____%O: 128.00 g/mol x 100 = 41.18% 310.85 g/mol % O = _41.18% ___ How many grams of tin are in 250g of tin (IV) sulfate?250 g Sn(SO4)2 x 118.71 g/mol Sn = 95.48 g Sn 310.85 g/mol Sn(SO4)2 MOLE CONVERSIONS Moles/Mass, Moles/Particles, Moles/Volume1016635191770Review: Find the molar mass of each compound below: a. diphosphorus pentoxideP2O5___141.94 g/molcopper (II) sulfateCuSO4___159.62 g/molcalcium cyanideCa(CN)2____92.12 g/mol___aluminum nitrateAl(NO3)3_____213.01 g/mol__MOLE/MASS and MASS/MOLE/PARTICLE CONVERSIONS USING MOLAR MASS298640562865Conversion Factors1 mole = 6.02 x 1023 molecules 1 mole = 6.02 x 1023 formula units1 mole = molar mass in grams00Conversion Factors1 mole = 6.02 x 1023 molecules 1 mole = 6.02 x 1023 formula units1 mole = molar mass in grams-16002062865STEP 1: Write the correct formulaSTEP 2: Determine the molar massSTEP 3: Use dimensional analysis to convert00STEP 1: Write the correct formulaSTEP 2: Determine the molar massSTEP 3: Use dimensional analysis to convertEXAMPLES: -161925140335Find the mass of 4.50 moles of diphosphorus pentoxide.______638.73 g______4.50 moles P2O5 141.94 g P2O51 mole P2O5244792516510000112395016510000How many moles is 250.0g of copper (II) sulfate?______1.57 mole CuSO4_______250.0 g CuSO4 1 mole CuSO4-857255080 159.62 g CuSO4 Find the mass of 0.545moles of calcium cyanide.______ 50.21 g Ca(CN)2________-3810014859000.545 moles Ca(CN)292.12 g Ca(CN)21 mole Ca(CN)24. How many molecules are in 110g of diphosphorus pentoxide?_______ 4.67 x 1023molecules_4552950106680002438400106680009810759715500110g P2O51 mole P2O56.02 x 1023 molecules P2O5-1619251333500141.94 g P2O51 mole P2O5YOU TRY SOME: 575g of sodium sulfate to moles___4.05 moles Na2(SO4) 0.025moles of phosphorus pentachloride to grams ____ 5.21 g P2Cl5______ 15.0g of iron(III) nitrate to moles________________ 8.02 x 1023 molecules of carbon disulfide to grams_______________MOLES AND VOLUME OF MATTER Avogadro’s Hypothesis: at Standard temperature and pressure, one mole of any gas will expand to fill 22.4 L STP: Standard temperature and pressure; 0 °C and 1 atmosphere0176530How many is a mole?1 mole = 6.02 X 1023 atoms, molecules, formula units, or particlesHow heavy is a mole?1 mole = molar mass in gramsHow much space does a mole occupy?1 mole = 22.4 L (for a gas at STP)00How many is a mole?1 mole = 6.02 X 1023 atoms, molecules, formula units, or particlesHow heavy is a mole?1 mole = molar mass in gramsHow much space does a mole occupy?1 mole = 22.4 L (for a gas at STP)YOU TRY SOME-Make the following mole conversions: 14.0L of nitrogen gas at STP to moles 2.5g of chlorine gas at STP to molecules 2.24 x 1025 atoms of neon at STP to liters 13.3L of fluorine gas at STP to gramsEMPIRICAL AND MOLECULAR FORMULAS85280587630Compare the Empirical Formula and Molecular Formula of each compound above.What is an Empirical Formula? a formula with the lowest whole number ratio of elements in a compound - simplified (CH2)What is a Molecular Formula? a chemical formula of a molecular compound that shows the kinds and number of atoms present in a molecule of a compound – NOT simplified (C2H4) and (C5H10)YOU TRY SOME: Given the following molecular formulas, determine the empirical formula of each compound.a. P4O10b. C6H12O6c. C3H6O P2O5 CH2O C3H6OIf we know the empirical formula and the molar mass of a compound, we can determine the molecular formula of that compound by finding the ratio of the given molecular molar mass to the molar mass of the empirical formula. For example: Given the empirical formula CH and the molar mass 78g/mol, find the molecular formula.STEP 1: Find molar mass of the empirical formula: 12 g/mol + 1 g/mol = 13 g/molSTEP 2: Divide the given molecular molar mass by the empirical molar mass (from step 1): 78 g/mol = 6 13 g/molSTEP 3: Round your answer in Step 2 to the nearest whole number (this is the ratio): 6STEP 4: Multiple all subscripts in the empirical formula by the ratio from step 3: C6H6YOU TRY SOME: Determine the molecular formula of each compound The empirical formula of a compound is CH2O. Its molar mass is 360g/mol. Find its molecular formula.CH2O = 12 + 2(1) + 16 = 30 g/1 mol360 / 30 = 12 C12H24O12 The empirical formula of a compound is P2O5. Its molar mass is 284g/mol. Find its molecular formula.P2O5 = 2(31.0) + 5(16) = 142 g/1 mol284 / 142 = 2 P4O103. The empirical formula of a compound is CH2O. Its molar mass is 180g/mol. Find its molecular formula.CH2O = 12 + 2(1) + 16 = 30 g/1 mol180 / 30 = 6 C6H12O64. The empirical formula of a compound is HO . Its molar mass is 34g/mol. Find its molecular formula.HO = 1+ 16 = 17 g/1 mol34 / 17 = 6 H2O2HW#1: REVIEW- NAMING AND MOLE CONVERSIONSNAMEIonic or Covalent?FORMULAPotassium permanganateCovalentKMnO4Tricarbon octahydrideIonicC3H8Iron (III) carbonateIonicFe2(CO3)3Tetraphosphorus decoxideCovalentP4O10Aluminum sulfateIonicAl2(SO4)3Copper (II) sulfiteIonicCu(SO3)Magnesium iodideIonicMgI2Tetraphosphorus decoxideCovalentP4O10Mole conversions: (you may want to go back and look at your previous notes on mole conversions)a. Find the mass of 1.25 moles of K (mole → g)1.25 mol K x 39.1 g K = 48.88 g K 1 mol Kb. How many moles is 2.50g of oxygen? (hint: oxygen is diatomic) (g → mole)2.50 g O2 x 1 mol O2 = 0.078 mol O2 32.0 g O2c. Find the number of atoms in 5.45 g of phosphorus. (g → mole → atoms)5.45 g P x 1 mol P x 6.02 x 1023 atoms = 1.09 x 1023 atoms 30.1 g 1 molHW #2 MOLAR MASS/ PER CENT COMPOSITION1. Label each compound below as ionic or molecular. Write the formula and determine the molar mass of each POUNDIONIC/MOLECULAR?FORMULAMOLAR MASSBarium sulfateIBaSO4233.39 g/molTricarbonoctahydrideMC3H844.11 g/molIron (III) carbonateIFe2(CO3)3291.73 g/molTetraphosphorusdecoxideMP4O10283.88 g/molStrontium phosphateISr3(PO4)2452.8 g/mol2. Find the per cent composition of iron (III) carbonate.% Fe = 38.28% % C = 12.35%% O = 49.36% How many grams of iron are in 125g of iron (III) carbonate?125 g Fe2(CO3)3 x 111.6 g/mol Fe_______ = 47.88 g Fe 291.73 g/mol Fe2(CO3)3 What mass of iron(III) carbonate contains 10.0g of iron?10.0 g Fe x 291.73 g Fe2(CO3)3 = 26.14 g Fe2(CO3)3 111.6 g FeHW #3 PERCENT COMPOSITION Determine the percent composition of each of the compounds below:KMnO4% K = 24.74% Mn = 34.76% O = 40.50HCl% H = 2.77% Cl = 97.23Mg(NO3)2% Mg = 16.39% N = 18.89% O = 64.72(NH4)3PO4% N = 28.19% H = 8.13% P = 20.77% O = 42.92Al2(SO4)3% Al = 15.77% S = 28.12% O = 56.11 Solve the following problems:How many grams of oxygen can be produced from the decomposition of 100g of KClO3?K: 1 x 39.10 g/mol = 39.10 g/mol 100 g KClO3 x 48.00 g/mol O = 39.17 g O Cl: 1 x 35.45 g/mol = 35.45 g/mol 122.55 g/mol KClO3O: 3 x 16.00 g/mol = 48.00 g/molHow much iron can be recovered from 25.0g of Fe2O3? 17.49 gHow much silver can be produced from 125g of Ag2S? 108.82 gHW #4 CONVERSION PRACTICE1. Write the formula for each compound below. Then determine its molar mass.FORMULAMOLAR MASScalcium sulfateCaSO4136.14 g/molaluminum cyanideAl (CN)3105.04 g/molphosphorus triiodidePI3411.67 g/mol2. Use the molar masses you found in Part 1 to make the following conversions:a. Find the mass of 1.25moles of calcium sulfate.170.19 g CaSO4b. How many moles is 2.50g of aluminum cyanide?0.024 mol Al (CN)3c. Find the mass of 0.750moles of phosphorus triiodide.308.75 g PI3d. How many molecules is 12.50g of phosphorus triiodide 1.83 x 1022 molecules PI3e. Find the number of formula units in 10.0g of calcium sulfate.4.42 x 1022 formula units CaSO4HW #5 MIXED MOLE CONVERSION PRACTICE Convert the quantities below to moles: 14.0g of lead ___0.068 mol Pb ________ b. 20.5g of lithium hydroxide ______0.86 mol LiOH _____ c. 14.0L of oxygen gas at STP _____0.625 mol O2________ d. 15.0g of calcium nitrate ____2. Find the mass of:a. 10.0L of hydrogen gas at STP______0.90 g H2________b. 4.5 moles of sodium sulfate_______________________c. 3.21 x 1022 molecules of carbon tetrachloride_____8.20 g CCl4________ ................
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