Chapter 6 - Chemical Composition
Determine the number of moles of O in 1.7 moles of CaCO3. Given: 1.7 moles CaCO3 Find: Moles of O
3 mol O = 1 mol CaCO3 1.7 mol CaCO3 X 3 mol O/1 mol CaCO3 = 5.1 mol O
Without doing any calculations, determine which sample contains the most atoms. a) one gram of cobalt b) one gram of carbon c) one gram of lead
Cobalt molar mass = 58.693 g Carbon molar mass = 12.011 g Lead molar mass = 207.98 g b) one gram of carbon
What is a molecular formula?
A molecular formula is a formula for a compound that gives the specific number of each type of atom in a molecule.
A compound containing carbon and hydrogen has a molar mass of 56.11 g/mol and an empirical formula of CH2. Determine its molecular formula.
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Empirical Molar Mass = 12.011 g + 2(1.0079) = 14.0268 g/mol n = Molar Mass/Emp Molar Mass = 4 Molecular Formula = 4 X CH2 = C4H8
What is the mass of 1 mol of atoms?
The mass of 1 mol of atoms is its molar mass. The value of an element's molar mass in grams per mole is numerically equal to the element's atomic mass in mass units.
What is the equation of mass percent?
Mass Percent = Mass of C in a sample of a compound/ Mass of the sample of the compound X 100%
A silver ring contains 1.1 X 10^22 silver atoms. How many moles of silver are in the ring? Given: 1.1 X 10^22 Ag atoms Find: Moles of Ag
1 mol Ag = 6.022 X 10^23 Ag atoms 1.1 X 10^22 Ag atoms X 1 mol Ag/6.022 X 10^10 Ag atoms = .01826 mol Ag = 1.8 X 10^-2
How many gold atoms are in a pure gold ring containing 8.83 X 10^-2 mol Au. Given: 8.83 X 10^-2 mol Au Find: Number of Au atoms
1 mol Au = 6.022 X 10^23 Au atoms 8.83 X 10^-2 mol Au X 6.022 X 10^23 Au atoms/1 mol Au = 5.32 X 10^22 Au Atoms.
Calculate the mass percent of Cl in freon-114 (C2Cl4F2). Given: C2Cl4F2 Find: Mass % of Cl
1 mol C2Cl4F2 = 4 mol of Cl Molar Mass of Cl = 35.453 g Molar Mass of C2Cl4F2 = 203.8302 g 4 (35.453) = 141.812 g Mass % Cl = 4 X Molar Mass Cl/Molar Mass of C2Cl4F2 X 100% 141.812 g/203.8302 = .6958 X 100% = 69.58%
Acetic Acid (HC2H3O2) is the active ingredient in vinegar. Calculate the mass percent composition of O in acetic acid. Given: HC2H3O2 Find: Mass % of O
1 mol HC2H3O2 = 2 mol of O Molar Mass of O = 15.999 Molar Mass of HC2H3O2 = 60.0520 g/mol Mass % of O = 2 X Molar Mass of Cl/Molar Mass of HC2HO2 X 100% 2 X 15.999/60.0520 = 0.5513 = 53.28%
Convert 3.5 mol of Helium (He) to a number of Helium (He) atoms. Given: 3.5 mol He Find: Number of He Atoms
1 mol He = 6.022 X 10^23 He atoms 3.5 mol He X 6.022 X 10^23He Atoms/1 mol He = 2.1 X 10^24 He Atoms
How many carbon atoms are in a diamond (pure carbon) with a mass of 38 mg? Given: 38 mg Find: C atoms
1 mol carbon = 12.011 g 1 mol carbon = 6.022 X 10^23 C atoms 38 mg X 1 g/1000 mg = .0038 g .0038 g C X 1 mol C/12.011 g C = .003163867 mol C .003163767 mol C X 6.022 X 10^23 C atoms/1 mol C = 1.9 X 10^21 C atoms
Determine the number of moles of O in 1.4 mol of H2SO4. Given: :1.4 mol H2SO4 Find: mol of O
1 mol of H2SO4 = 4 mol O 1.4 mol H2SO4 X 4 mol O/1 mol H2SO4 = 5.6 mol O
How many mercury (Hg) atoms are in 5.8 mol of mercury? Given: 5.8 mol of Hg Find: Number of Hg
1 mol of Hg = 6.022 X 10^23 Hg atoms 5.8 mol of Hg x 6.022 X 10^23 Hg atoms/1 mol Hg = 3.5 X 10^24 Hg atoms
How many titanium atoms are in a pure titanium bicycle frame with a mass of 1.28 kg? Given: 1.28 kg Ti Find: Ti atoms
1 mol of Ti = 47.867 g 1 mol of Ti = 6.022 X 10^23 Ti atoms 1.28 kg X 1000 g/1 kg = 1280 g 1280 g X 1 mol Ti/47.867 g = 26.7407 mol Ti 26.7407 mol Ti X 6.022 X 10^23 Ti atoms/1 mol Ti = 1.61 X 10^25 Ti Atoms
Calculate the number of moles of O in 18 moles of CO2. Given: 18 moles CO2 Find: Moles of O
2 mol O = mol CO2 18 moles CO2 X 2 moles O/1 mole CO2 = 36 mol O
In small amounts, the fluoride ion (often consumes as NaF) prevents tooth decay. According to the American Dental Assoc, an adult female should only consume 3.0 mg of fluorine per day. Calculate the amount of sodium fluoride (45.24 %F) that a woman should consume to get the recommended amount of fluorine. Given: 3.0 mg F; Mass % F = 45.24%
3.0 mg F X 1 g/1000 g = 0.003 g F 45.24 g F = 100 g NaFl 0.003 g F X 100 g NaFl/45.24 g F = 0.00663 g 6.6 mg NaFl
A certain type of nail weighs 0.50 lb per dozen. How many nails are contained in 3.5 lb of these nails? a) 84 b) 21 c) 0.58 d) 12
3.5 lb X 12 nails/0.50 lb = 84 nails a) 84
Copper (II) fluoride contains 37.42% F by mass. Use this percentage to calculate the mass of fluoride in grams contained in 28.5 g of Copper (II) Fluoride Given: 28.5 CuF; Mass % of F = 37.42% Find: F in grams
37.42 grams F = 100 grams CuF 28.5 g CuF X 37.42 g F/100 g CuF = 10.7 g F
The FDA recommends that adults consume less than 2.4 g of sodium per day. How many grams of sodium chloride can you consume and still be within FDA guidelines? Sodium Chloride is 39% sodium by mass. Given: 2.4 g Na; NaCl = 39% Na by mass Find: NaCl grams
39 g Na = 100 g NaCl 2.4 g Na X 100 g NaCl/39 g Na = 6.2 g NaCl
If a woman consumes 22 grams of sodium chloride, how much sodium does she consume? Sodium chloride is 39% sodium by mass. Given: 22 g NaCl; Mass % Na = 39% Find: Na in Grams
39 grams Na = 100 grams NaCl 22 g NaCl X 39 g Na/100 grams NaCl = 8.6 g Na
What is the formula mass of CO2?
Atomic Mass C = 12.011 amu Atomic Mass O = 15.999 amu 12.011 + 2(15.999) = 44.009 g/mol
Calculate the mass (in grams) of 1.75 mol of water. Given: 1.75 mol H2O Find: Mass of H2O
Atomic Mass H2O = 2(1.0079) + 15.999 = 18.0148 g/mol 1 mol H2O = 18.0148 g H2O 1.75 mol H2O X 18.0148 g H2O/1 mol H2O = 31.5259 g H2O = 31.5 g H2O
How many H2O molecules are in a sample of water with a mass of 3.64g Given: 3.64 g H2O Find: H2O molecules
Atomic Mass H2O = 2(1.0079) + 15.999 = 18.0148 g/mol 1 mol H2O = 18.0148 g H2O 1 mol H2O = 6.022 X 10^23 H2O molecules 3.64 g H2O X 1 mol H2O/18.0148 g H2O = 0.202056 mol H2O 0.020056 mol H2O X 6.022 X 10^23 H2O molc/1 mol H2O= 1.2167 X 10^23 H2O molecules = 1.22 X 10^23 H2O molecules
How many aluminum (Al) atoms are in an aluminum can with a mass of 16.2 g? Given: 16.2 g Al Find: Al atoms
Atomic Mass of Al = 26.982 g 1 mol Al = 26.982 g Al 1 mol Al = 6.022 X 10^23 Al atoms 16.2 g Al X 1 mol Al/26.982 g Al = .6004002 mol Al .6004002 mol Al X 6.022 X 10^23 mol atoms/1 mol Al = 3.62 X 10^23 Al atoms
How many aluminum atoms are in 3.78 g of aluminum? Given: 3.78 g Al Find: Al Atoms
Atomic Mass of Al = 26.982 g 1 mol Al = 26.982 g Al 1 mol Al = 6.022 X 10^23 Al atoms 3.78 g Al X 1 mol Al/26.982 g Al = .1400933 mol Al .1400933 mol Al X 6.022 X 10^23 Al Atoms/1 mol Al = 8.44 X 10^22 Al atoms
Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3. Given: 7.20 g Al2(SO4)3 Find: O in grams
Atomic Mass of Al2(SO4)3 = 342.1509 grams Atomic Mass of O = 15.999 grams 1 mol Al2(SO4)3 = 12 mol O 1 mol Al2(SO4)3 = 342.1059 grams 1 mol O = 15.999 grams 7.20 g Al2(SO4)3 X 1 mol Al2(SO4)3/342.1059 grams = 0.02104 mol Al2(SO4)3 X 12 mol O/1 mol Al2(SO4)3 = 0.252 mol O X 15.999 g O/1 mol O = 4.04 grams O
Calculate the number of moles of carbon in a 0.58 g diamond (pure carbon). Given: 0.58 g C Find: Moles of C
Atomic Mass of C = 12.011 g 1 mol of C = 12.011 g of C 0.58 g C X 1 mol of C/12.011 g C = .048 mol of C = 4.8 X 10^-2 mol C
Calculate the number of carbon atoms in 0.58 g diamond. Given: 0.58 g C Find: Number of Carbon Atoms
Atomic Mass of C = 12.011 g 1 mol of C = 12.011 g of C 1 mol of C = 6.022 X 10^23 C atoms 0.58 g C X 1 mol C/12.011 g C = .048289 mol C .048289 mol C X 6.022 X 10^23 C Atoms/1 mol C = 2.9 X 10^22 C atoms
Carvone (C10H14O) is the main component of spearmint oil. It has a pleasant aroma and mint flavor. Carvone is added to chewing gum, liqueurs, soaps and perfumes. Calculate the mass of carbon in 55.4 g of carvone. Given: 55.4 g C10H14O Find: C in Grams
Atomic Mass of C10H14O = 150.2196 g Atomic Mass of C = 12.011 g 10 mol C = 1 mol C10H14O 1 mol C = 12.011 grams 1 mol C10H14O = 150.2196 g 55.4 g C10H14O X 1 mol C10H14O/150.2196 C10H14O g = 0.3687 mol C10H14O X 10 mol C/1 mol C10H141O = 3.687 mol C X 12.011 g C/1 mol C = 44.3 C
A mothball, compose of naphthalene (C10H8) has a mass of 1.32 g. How many naphthalene molecules does it contain? Given: 1.32 g C10H8 Find: C10H8 molecules
Atomic Mass of C10H8 = 10(12.011) + 8(1.0079) = 128.1732 g/mol 1 mol C10H8 = 128.1732 g C10H8 1 mol C10H8 = 6.022 X 10^23 molecules of C10H8 1.32 g C10H8 X 1 mol C10H8/128.1732 g C10H8 = 0.01029 mol C10H8 0.01029 mol C10H8 X 6.022 X 10^23 molc C10H8/1 mol C10H8 = 6.20 X 10^21 C10H8 molecules
Find the number of CO2 molecules in a sample of dry ice (solid CO2) with a mass of 22.5 g. Given: 22.5 g CO2 Find: CO2 molecules
Atomic Mass of CO2 = 12.011 + 2(15.999) = 44.009 1 mol CO2 = 44.009 g 1 mol CO2 = 6.022 X 10^23 CO2 molecules 22.5 g CO2 X 1 mol CO2/44.009 g CO2 = 0.511259 mol CO2 0.511259 mol CO2 X 6.022 X10^23 CO2 molecules/1 molCO2 = 3.08 X 10^23 molecules CO2
Find the number of moles in a 22.5 g sample of dry ice (solid CO2). Given: 22.5 g CO2 Find: CO2 moles
Atomic Mass of CO2 = 44.0009 1 mol CO2 = 44.009 g 22.5 g X 1 mol CO2/44.009 g = .551259 mol CO2 = .511 mol CO2
How many grams of Cl are in 38.0 g of each sample of chlorofluorocarbons (CFCs)? a) CF2Cl2 b) CFCl3 c) C2F3Cl3 d) CF3Cl
Atomic Mass of Cl = 35.453 grams 1 mol Cl = 35.453 grams a) CF2Cl2 = Atomic Mass = 120.9135 grams 1 mol CF2Cl2 = 120.9135 grams 1 mol CF2Cl2 = 2 mol Cl 38.0 g CF2Cl2 X 1 mol CF2Cl2/120.9135 g CF2Cl2 = 0.3142 mol CF2Cl2 X 2 mol Cl/1 mol CF2Cl2 = 0.629 mol Cl 0.629 mol Cl X 35.453 g Cl/1 mol Cl = 22.3 g Cl b) CFCl3 = Atomic Mass = 137.3681 grams 1 mol CFCl3 = 137.3681 grams 1 mol CFCl3 = 3 mol Cl 38.0 g CFCl X 1 mol CFCl3/137.3681 g CFCl3 = 0.2766 mol CFCl3 X 3 mol Cl/1 mol CFCl3 = 0.82988 mol Cl 0.82988 mol Cl X 35.453 g Cl/1 mol Cl = 29.4 g Cl c) C2F3Cl3 = Atomic Mass = 187.3756 grams 1 mol C2F3Cl3 = 187.3756 grams 1 mol C2F3Cl3 = 3 mol Cl 38.0 g C2F3Cl3 X 1 mol C2F3Cl3/187.3756 g = 0.2028 mol C2F3Cl3 X 3 mol Cl/1 mol C2F3Cl3 = 0.6084 mol Cl X 35.453 g Cl/1 mol Cl = 21.6 g Cl d) CF3Cl = Atomic Mass = 104.4589 grams 1 mole CF3Cl = 104.4589 gram CF3Cl 1 mol CF3Cl = 1 mol Cl 38.0 g CF3Cl X 1 mol CF3Cl/104.4589 g = 0.3637 mol CF3Cl 0.3638 mol CF3Cl X 1 mol Cl/1 mol CF3Cl = 0.3638 mol Cl 0.3638 mol Cl X 35.453 Cl/1 mol Cl = 12.9 g Cl
Calculate the number of moles of NO2 in 1.18 g of NO2. Given: 1.18 g NO2 Find: NO2 Moles
Atomic Mass of NO2 = 14.007 + 2(15.999) = 46.005 g/mol 1 mol NO2 = 46.005 g 1.18 g NO2 X 1 mol NO2/46.005 g NO2 = .02564 mol NO2 2.56 X 10^-2 mol NO2
What is the mass of 4.78 X 10^24 NO2 molecules. Given: 4.78 X 10^24 NO2 molecules Find: Mass of NO2
Atomic Mass of NO2 = 14.007 + 2(15.999) = 46.005 g/mol 1 mol NO2 = 46.005 g NO2 1 mol NO2 = 6.022 X 10^23 NO2 molecules 4.78 X 10^24 NO2 molc X 1 mol NO2/6.022 X 10^23 molc = 7.93756 mol NO2 7.93756 mol NO2 X 46.005 g NO2/1 mol NO2 = 365 g NO2
Determine the mass of Sodium (Na) in 15g of NaCl. Given: 15 g NaCl Find: Na in grams
Atomic Mass of Na = 22.990 grams Atomic Mass of NaCl = 58.443 grams 1 mol Na = 1 mol NaCl 1 mol Na = 22.990 grams 1 mol NaCl = 58.443 grams 15 g NaCl X 1 mol NaCl/58.443 grams = 0.25666 mol NaCl 0.25666 mol NaCl X 1 mol Na/1 mol NaCl = 0.25666 mol Na 0.25666 mol Na X 22.990 g Na/1 mol Na = 5.9 g Na
Determine the mass of oxygen in a 5.8 g sample of sodium bicarbonate (NaHCO3). Given: 5.8 g NaHCO3 Find: O in grams
Atomic Mass of NaHCO3 = 84.0059 grams Atomic Mass of O = 15.999 grams 1 mol NaHCO3 = 3 mol O 1 mol NaCO3 = 84.0059 g 1 mol O = 15.999 g 5.8 g NaHCO3 X 1 mol NaCO3/84.0059 g NaCO3 = 0.069 mol NaCO3 X 3 mol O/1 mol NaCO3 = 0.207 mol O 0.207 mol O X 15.999 g/1 mol O = 3.3 grams O
Calculate the number of moles of sulfur in 57.8 g of Sulfur. Given: 57.8 g S Find: Mole of S
Atomic Mass of S = 32.065 g 1 mol of S = 32.065 g 57.8 g S X 1 mol of S/32.065 g = 1.80 mol S
Calculate the mass of 1.23 X 10^24 He atoms. Given: 1.23 X 10^24 He atoms Find: Mass of He
Atomic mass of He = 4.0026 g 1 mol of He = 4.0026 g 1 mol of He = 6.022 X 10^23 He atoms 1.23 X 10^24 He atoms X 1 mol He/6.022 X 10^23 He atoms= 2.0425 mol He 2.0425 mol He X 4.0026 g He/1 mol He = 8.17 g He
Compound A has a molar mass of 100 g/mol and Compound B has a molar mass of 200 g/mol. If you have samples of equal mass of both compounds, which sample contains the greater number of molecules.
Compound A. A would have the greatest number of molecules. Sample A has a lower molar mass than sample B, so a given mass of Sample A has more moles and therefore more molecules than the same mass of sample B.
What is empirical formula mass?
Empirical formula molar mass is the sum of the molar masses of all of the atoms in an empirical formula.
A 0.358 g sample of Chromium reacts with Oxygen to form 0.523 g of metal oxide. What is the mass percent of chromium? Given: 0.358 g Cr; 0.523 g metal oxide Find: Mass Percent of Cr
Mass Percent of Cr = 0.358 g Cr/0.523 g metal oxide = 0.6845 X 100% = 68.5 %
What is mass percent composition (or mass percent)?
Mass percent composition or mass percent is the percentage by mass of each element in a compound.
What is another formula for the mass percent of an element?
Mass percent of element X = Mass of Element X in 1 mol of compound/Mass of 1 mol of compound X 100%
Butane is a compound containing carbon and hydrogen used as a fuel in butane lighters. Its empirical formula is C2H5, and its molar mass is 58.12 g/mol. Find its molecular formula.
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Empirical Molar Mass = 2(12.011) + 5(1.0079) = 29.0615 g/mol n = Molar Mass/Emp Molar Mass = 58.12/29.0615 = 2 Molecular Formula = C2H5 X 2 = C4H10
Naphthalene is a compound containing carbon and hydrogen that is used in mothballs. It empirical formula is C5H4 and its molar mass is 128.16 g/mol. What is it molecular formula. Given: Empirical Formula = C5H4; C5H4 molar mass = 128.16 Find: Molecular Formula
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Empirical formula molar mass = 5(12.011 g) + 4(1.0079 g) = 64.0866 g/mol n = Molar Mass/Empirical Formula Mass = 128.16/64.0866 = 2 Molecular Formula = C5H4 X 2 = C10H8
A compound with the following mass percent has a molar mass of 60.10 g/mol. Find its molecular formula. C = 39.97% H = 13.41% N = 46.62%
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Molar Mass of N = 14.007 g 39.97 g C X 1 mol C/12.011 g C = 3.3277 mol of C 13.41 g H X 1 mol H/1.0079 g H = 13.3048 mol H 46.62 g N X 1 mol N/14.007 g = 3.3283 mol N 3.3277/3.3277 = 1; 13.3048/3.3277 = 3.998; 3.3283/3.3277 = 1 Empirical Formula = CH4N 12.011g + 4(1.0079g) + 14.007g = 30.0496 mol/g n = Molar Mass/Emp Molar Mass = 60.10/30.0496 = 2 Molecular Formula = 2 X CH4N = C2H8N2
The rotten smell of decaying animal carcass is partially due to a nitrogen-containing compound called putrescence. Elemental analysis of the putrescence indicates that it consists of 54.5% C 13.73% H 31.77% N Calculate the empirical formula of putrescence
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Molar Mass of N = 14.007 g 54.5 g C X 1 mol C/12.011 g = 4.538 moles C 13.73 g H X 1 mol H/1.0079 g = 13.622 moles H 31.77 g N X 1 mol N/14.007 g = 2.268 moleN 4.538/2.268 = 2.00; 13.622/2.268 = 6.00; 2.268/2.268 = 1 C2H6N
A sample of a compound is decomposed in the laboratory and produces 165 g of carbon, 27.8 g of hydrogen, ,and 220.2 g O. Calculate the empirical formula of the compound.
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Molar Mass of O = 15.999 g 165 g C X 1 mol C/12.011 = 13.737 mol C 27.8 g H X 1 mol H/1.0079 g = 27.58 mol H 220.2 g O X 1 mol O/15.999 g = 13.763 mol O 13.737/13.763 = 0.998; 27.58/13.763 = 2.00; 13.763/13.763 = 1 CH2O
A laboratory analysis of aspirin determines the following mass percent composition. C - 60.00% H - 4.48% O - 35.53% Find the empirical formula.
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Molar Mass of O = 15.999 g 60.00 g C X 1 mol C/12.011 g = 4.996 mol C 4.48 g H X 1 mol H/1.0079 g = 4.445 mol H 35.53 g O X 1 mol O/15.999 g = 2.221 mol O C_4.996H_4.445O_2.221 - divide by smallest subscript (mol#) 4.996/2.221 = 2.25; 4.445/2.221= 2.001; 2.221/2.221 = 1 C_2.25H_2O_1 X 4 = C9H8O4
Ibuprofen has the mass percent composition: C 75.69%; H 8.80%; O 15.51%. Calculate the empirical formula of the ibuprofen.
Molar Mass C = 12.011 g Molar Mass H = 1.0079 g Molar Mass of O = 15.999 g 75.69 g C X 1 mol C/12.011 = 6.302 mol C 8.80 g H X 1 mol H/1.0079 g = 8.731 mol H 15.51 g O X 1 mol O/15.999 = 0.969 mol O 6.302/0.969 = 6.5; 8.731/.969 = 9.01; .969/.969 = 1 C_6.5H_.9O_1 C13H18O2
Calculate the mass percent composition of nitrogen in each compound. a) N2O b) NO c) NO2 d) N2O5
Molar Mass of N = 14.007 g a) N2O - Molar Mass = 44.0128 g 1 mol N2O = 2 mol N Mass % of N = 2 X 14.007/44.0128 = 0.6365 X 100% = 63.65% b) NO = Molar Mass = 30.0061 g 1 mol NO = 1 mol N Mass % of N = 1 X 14.007/30.0061 = 0.4668 X 100% = 46.68% c) NO2 = Molar Mass = 46.0055 g 1 mol NO2 = 1 mol N Mass % of N = 1 X 14.007/46.0055 = 0.3045 X 100% = 30.45% d) N2O5 = Molar Mass = 108.0104 g 1 mol N2O5 = 2 mol N 2 X 14.007 / 108.0104 = 0.2593 X 100% = 25.94%
Decompose a compound containing nitrogen and oxygen in the laboratory and produce 24.5 g of nitrogen and 70.0 of oxygen Calculate the empirical formula of the compound. Given: 24.5 g N; 70.0 O Find: Empirical Formula of Compound
Molar Mass of N = 14.007 g Molar Mass of O = 15.999 g 24.5 g N X 1 mol N/14.007 g = 1.75 mol N 70.0 g O X 1 mol O/15.999 g = 4.38 mol O N_1.75O_4.38 - divide each by smallest subscripts (mol #) 1.75/1.75 = 1; 4.38/1.75 = 2.50 N1O2.5 - Make each subscript a whole number N2O2.5 X 2 = N2O5.
A 1.45 g sample of phosphorus burns in air and forms 2.57 g of a phosphorus oxide. Calculate the empirical formula of the oxide.
Molar Mass of P = 30.974 g Molar Mass of O = 15.999 g Mass of O = 2.57 g - 1.45 g = 1.12 g O 1.45 g P X 1 mol P/30.974 g P = 0.0468 mol P 1.12 g O X 1 mol O/15.999 g O = 0.0700 mol O 0.0468/0.0468 = 1; .0700/.0468 = 1.49; PO_1.5 X 2 = P2O3
A 3.24 g sample of titanium reacts with oxygen to form 5.4 g of the metal oxide. What is the empirical formula? Given: 3.25 g Ti; 5.4 g Metal oxide Find: Empirical Formula
Molar Mass of Ti = 47.867 g Molar Mass of O = 15.999 g Mass O = Mass of Metal Oxide - Mass of Ti Mass O = 5.4 g - 3.24 g = 2.16 g 3.24 g Ti X 1 mol Ti/47.867 g = 0.06768 mol Ti 2.16 g O X 1 mol O/15.999 g = 0.1350 mol O 0.06768/0.06768 = 1; 0.1350/0.6768 = 1.994 TiO2
Determine the number of moles of C in each sample. a) 2.5 mol CH4 b) 0.155 mol C2H6 c) 5.67 mol C4H10 d) 25.1 mol C8H18
a) 2.5 mol CH4 1 mol of CH4 = 1 mol C 2.5 mol CH4 X 1 mol C/1mol CH4 = 2.5 mol C. b) 0.115 mol C2H6 1 mol C2H6 = 2 mol C 0.115 mol C2H6 X 2 mol C/1 mol C2H6 = 0.23 mol C c) 5.67 mol C4H10 1 mol C4H10 = 4 mol C 5.67 mol C4H10 X 4 mol C/1 mol C4H10 = 22.68 mol C d) 25.1 mol C8H18 1 mol C8H18 = 8 mol C 25.1 mol C8H10 X 8 mol c/1 mol C8H18 = 200.8 mol C
Which statement is always true for samples of atomic elements, regardless of the type of element in the samples? a) If two samples of different elements contain the same number of atoms, the contain the same number of moles. b) If two samples of different elements have the same mass, they contain the same number of moles. c) If two samples of different elements have the same mass, the contain the same number of atoms.
a) If two samples of different elements contain the same number of atoms, the contain the same number of moles.
What is an empirical formula?
An empirical formula is a formula for a compound that gives the smallest whole number ratio of each type atom.
What is Avogadro's Number?
Avogardro's number is the number of entities in a mole, 6.022 X 10^23. It was named after Amadeo Avogadro.
What does sodium do for our bodies?
Sodium is involved in the regulation of body fluids. Eating too much sodium can lead to high blood pressure. High blood pressure increases the risk of stroke and heart attack.
What is the numerical value of the mole?
The numerical value of the mole is defined as being equal to the number of atoms in exactly 12 g of pure carbon-12.
What is the value of a mole (mol)?
The value of a mole (mol) is 6.022 X 10^23. 1 mole = 6.022 X 10^23 atoms.
Which compound has the highest mass percent of O? (You should not have to perform any detailed calculations to answer this question). a) CrO b) CrO2 c) Cr2O2
b) CrO2 This compound has the highest ratio of oxygen atoms to chromium atoms and therefore has the greatest mass percent of oxygen.
Without doing any detailed calculations, determine which sample contains the most fluorine atoms. a) 25 g of HF b) 1.5 mol of CH3F c) 1.0 mol of F2
c) 1.0 mol of F2 1.0 mol F2 contains 2.0 mol of F atoms. The the other two options each contain less than two moles of F atoms
A compound containing nitrogen and oxygen is decomposed in the laboratory and produces 1.78 g of nitrogen and 4.05 g of oxygen. Calculate the empirical formula of the compound.
Molar Mass of O = 15.999 g Molar Mass of N = 14.007 g 1.78 g N X 1 mol N/14.007 g = 0.127 mol N 4.05 g O X 1 mol O/15.999 g = 0.253 mol O 0.127/0.127 = 1; 0.253/0.127 = 2 NO2
Determine the number of moles of molecules (or formula units) in each sample. a) 38.2 g sodium chloride b) 36.5 g nitrogen monoxide c) 4.25 kg carbon dioxide d) 2.71 mg carbon tetrachloride
a) 38.2 g sodium chloride - NaCl Atomic Mass of NaCl = 22.990 + 35.453 = 58.443 g/mol 1 mol NaCl = 58.453 g NaCl 38.2 g NaCl X 1 mol NaCl/58.453 g NaCl = 0.6535 mol NaCl = 6.54 X 10^-1 mol NaCl b) 36.5 g nitrogen monoxide - NO Atomic Mass of NO = 14.007 + 15.999 = 30.006 g/mol 1 mol NO = 30.006 mol NO 36.5 g NO X 1 mol NO/30.006 mol NO = 1.22 mol NO c) 4.25 kg carbon dioxide = CO2 Atomic Mass of CO2 = 12.011 + 2(15.999) = 44.009 1 mol CO2 = 44.009 g CO2 4.25 kg X 1000 g/1 kg = 4250 g CO2 4250 g CO2 X 1 mol CO2/44.009 g CO2 = 96.6 mol CO2 d) 2.71 mg carbon tetrachloride - CCl4 Atomic Mass of CCl4 = 12.011 + 4(35.453) = 153.823 g/mol 1 mol CCl4 = 153.823 g CCl4 2.71 mg CCl4 X1 g/1000 mg = .00271 g CCl4 .002741 g CCl4 X 1 mol CCl4/153.823 g CCl4 = .000017618 mol ClCl4 = 1.76 X 10^-5 mol ClCl4
How many atoms are in each elemental sample. a) 16.9 g Sr b) 26.1 g Fe c) 8.55 g Bi d) 38.2 g P
a) Given: 16.9 g Sr; Find: Sr Atoms Atomic Mass Sr = 87.62 g 16.9 g Sr X 1 mol Sr/87.62 g Sr = .192878 mol Sr .192878 mol Sr X 6.022 X 10^23 Sr atoms/1 mol Sr = 1.16 X 10^23 Sr atoms b) Given: 26.1 g Fe: Find: Fe atoms Atomic Mass Fe = 55.845 g 26.1 g Fe X 1 mol Fe/55.845 g Fe = 0.46736 mol Fe 0.46736 mol Fe X 6.022 X 10^23 Fe atoms/1 mol Fe= 2.81 X 10^23 Fe atoms c) Given: 8.55 g Bi; Find Bi Atoms Atomic Mass of Bi = 208.98 g 8.55 g Bi X 1 mol Bi/208.98 g Bi = .040913 mol Bi 0.040913 mol Bi X 6.022 X 10^23 Bi atoms/1 mol Bi = 2.46 X 10^22 Bi atoms d) Given: 38.2 g P; Find: P Atoms Atomic Mass of P = 30.974 g 38.2 g P X 1 mol P/30.974 g P = 1.23329 mol P 1.23329 mol P X 6.022 X 10^23 P atoms/1 mol P = 7.43 X 10^23 P atoms
How many atoms are in each chemical sample? a) 3.4 mol Cu b) 9.7 X 10^-3 mol C c) 22.9 mol Hg d) 0.215 mol Na
a) Given: 3.4 mol Cu; Find: Cu Atoms 3.4 mol Cu X 6.022 X 10^23 Cu Atoms/1 mol Cu = 2.05 X 10^24 Cu atoms b) Given: 9.7 X 10^-3 mol C; Find: C Atoms 9.7 X 10^-3 mol C X 6.022 X 10^23 C atoms/1 mol C = 5.8 X 10^21 C atoms c) Given: 22.9 mol Hg; Find: Hg Atoms 22.9 mol Hg X 6.022 X 10^23 Hg Atoms/1 mol Hg = 1.38 X 10^25 Hg Atoms d) Given: 0.215 mol Na; Find: Na Atoms 0.215 mol Na X 6.022 X 10^23 Na Atoms/1 mol Na = 1.29 X 10^23 Na Atoms