CnHm + (n+ )O2(g) n CO2(g) + H2O(g)m 2

m 2

Fig. 3.4

Ascorbic acid ( Vitamin C ) - I contains only C , H , and O

• Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O.

• Calculate its Empirical formula!• C:

• H:

• Mass Oxygen =

Ascorbic acid ( Vitamin C ) - I contains only C , H , and O

• Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O.

• Calculate its Empirical formula!

• C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)

= 2.65 x 10-3 g C• H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)

= 2.95 x 10-4 g H

• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O = 3.54x10-3 g O

Vitamin C combustion - II• C =

• H =

• O =

• Divide each by smallest:

• C =• H = • O =

Vitamin C combustion - II• C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C• H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H• O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O

• Divide each by smallest (2.21 x 10-4 ):• C = 1.00 Multiply each by 3: C = 3.00 = 3.0• H = 1.32 H = 3.96 = 4.0• O = 1.00 O = 3.00 = 3.0

C3H4O3

Determining a Chemical Formula from Combustion Analysis - I

Problem: Erthrose (M = 120 g/mol) is an important chemical compound used often as a starting material in chemical

synthesis, and contains Carbon, Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded:

1.027 g CO2 and 0.4194 g H2O.

Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.

Determining a Chemical Formula from Combustion Analysis - II

Calculating the mass fractions of the elements:

Mass fraction of C in CO2 =

Mass fraction of H in H2O =

Calculating masses of C and H:

Mass of Element = mass of compound x mass fraction of element

Determining a Chemical Formula from Combustion Analysis - II

Calculating the mass fractions of the elements:

Mass fraction of C in CO2 = =

= = 0.2729 g C / 1 g CO2

Mass fraction of H in H2O = =

= = 0.1119 g H / 1 g H2O

Calculating masses of C and H

Mass of Element = mass of compound x mass fraction of element

mol C x M of Cmass of 1 mol CO2

1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2

mol H x M of Hmass of 1 mol H2O

2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

Determining a Chemical Formula from Combustion Analysis - III

Mass (g) of C =

Mass (g) of H =

Calculating the mass of O:

Calculating moles of each element: C = H = O =

Determining a Chemical Formula from Combustion Analysis - III

Mass (g) of C = 1.027 g CO2 x = 0.2803 g C

Mass (g) of H = 0.4194 g H2O x = 0.04693 g H

Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g OCalculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O

C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4

0.2729 g C 1 g CO2

0.1119 g H 1 g H2O

Fig. 3.5

MolesMoles

Molecules

Avogadro’s Number

Molecular Formula

Atoms

x6.022 x 1023

Chemical Equations

Reactants Products

Qualitative Information:

States of Matter: (s) solid(l) liquid(g) gaseous(aq) aqueous

2 H2 (g) + O2 (g) 2 H2O (g)But also Quantitative Information!

Fig. 3.6

Fig. 3.7

Balanced Equations

1 CH4 (g) + O2 (g) 1 CO2 (g) + H2O (g)

• Mass Balance (or Atom Balance)- same number of each element on each side of the equation:

(1) start with simplest element(2) progress to other elements(3) make all whole numbers(4) re-check atom balance

• Make charges balance. (Remove “spectator” ions.)1 CH4 (g) + 2 O2 (g) 1 CO2 (g) + 2 H2O (g)

1 CH4 (g) + O2 (g) 1 CO2 (g) + 2 H2O (g)

Ca2+ (aq) + 2 OH- (aq) Ca(OH)2 (s)+ Na+ + Na+

Information Contained in a Balanced Equation

Viewed in Reactants Productsterms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy

Molecules

Amount (mol)

Mass (amu)

Mass (g)

Total Mass (g)

Information Contained in a Balanced Equation

Viewed in Reactants Productsterms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy

Molecules 2 molecules of C2H6 + 7 molecules of O2 = 4 molecules of CO2 + 6 molecules of H2OAmount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O

Mass (amu) 60.14 amu C2H6 + 224.00 amu O2 = 176.04 amu CO2 + 108.10 amu H2O

Mass (g) 60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O

Total Mass (g) 284.14g = 284.14g

Balancing Chemical Equations - I

Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C6H14).Plan: Write the skeleton equation from the words into chemical compounds with blanks before each compound. begin the balance with the most complex compound first, and save oxygen until last!Solution: Begin with one unit of most complex compound:

Balance any elements that this forces:

Balancing Chemical Equations - I

Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C6H14).Plan: Write the skeleton equation from the words into chemical compounds with blanks before each compound. Begin the balance with the most complex compound first, and save oxygen until last!Solution:

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

Begin with one C6H14 molecule which says that we will get 6 CO2’s!

1 6

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

Next balance H atoms:

Balancing Chemical Equations - II

1 6

Balance O atoms last:

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

The H atoms in the hexane will end up as H2O, and we have 14 H atoms, and since each water molecule has two H atoms, we will geta total of 7 water molecules.

Balancing Chemical Equations - II

1 6 7

Since oxygen atoms only come as diatomic molecules (two O atoms, O2),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore multiply the hexane by 2, giving a total of 12 CO2 molecules, and 14 H2O molecules.

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy2 12 14

This now gives 12 O2 from the carbon dioxide, and 14 O atoms from thewater, which will be another 7 O2 molecules for a total of 12+7 =19 O2 !

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy2 12 1419

Fig. 3.8

Chemical Equation Calc - I

Reactants ProductsMolecules

Atoms (Molecules)Avogadro’sNumber

6.02 x 1023

Chemical Equation Calc - II

Reactants ProductsMolecules

Moles

MassMolecularWeight g/mol

Atoms (Molecules)Avogadro’sNumber

6.02 x 1023

Sample Problem: Calculating Reactants and Products in a Chemical Reaction - IProblem: Given the following chemical reaction between AluminumSulfide and water, if we are given 65.80 g of Al2S3: a) How many molesof water are required for the reaction? b) What mass of H2S & Al(OH)3

would be formed? Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g)

Plan: Calculate moles of Aluminum Sulfide using its molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it’s molecular weight.Solution: a) molar mass of Aluminum Sulfide =

moles Al2S3 =

Sample Problem: Calculating Reactants and Products in a Chemical Reaction - IProblem: Given the following chemical reaction between AluminumSulfide and water, if we are given 65.80 g of Al2S3: a) How many molesof water are required for the reaction? b) What mass of H2S & Al(OH)3

would be formed? Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g)

Plan: Calculate moles of Aluminum Sulfide using its molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it’s molecular weight.Solution: a) molar mass of Aluminum Sulfide = 150.17 g / mol

moles Al2S3 = = 0.4382 moles Al2S3

65.80 g Al2S3

150.17 g Al2S3/ mol Al2S3

Calculating Reactants and Products in a Chemical Reaction - II

a) cont. H2O: 0.4382 moles Al2S3 x

b)H2S: 0.4382 moles Al2S3 x

___ moles H2O1 mole Al2S3

___ moles H2S1 mole Al2S3

molar mass of H2S =

mass H2S =

Al(OH)3: 0.4382 moles Al2S3 x molar mass of Al(OH)3 =

mass Al(OH)3 =

Calculating Reactants and Products in a Chemical Reaction - II

a) cont. 0.4382 moles Al2S3 x = 2.629 moles H2O

b) 0.4382 moles Al2S3 x = 1.314 moles H2S

molar mass of H2S = 34.09 g / mol

mass H2S = 1.314 moles H2S x = 44.81 g H2S

0.4382 moles Al2S3 x = 0.8764 moles Al(OH)3

molar mass of Al(OH)3 = 78.00 g / mol mass Al(OH)3 = 0.8764 moles Al(OH)3 x = = 68.36 g Al(OH)3

6 moles H2O1 mole Al2S33 moles H2S

1 mole Al2S3

34.09 g H2S1 mole H2S

2 moles Al(OH)3

1 mole Al2S3

78.00 g Al(OH)3

1 mole Al(OH)3

Calculating the Amounts of Reactants and Products in a Reaction Sequence - I

Problem: Calcium Phosphate could be prepared in the following reaction sequence:

4 P4 (s) + 10 KClO3 (s) 4 P4O10 (s) + 10 KCl (s)

P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq)

2 H3PO4 (aq) + 3 Ca(OH)2 (aq) 6 H2O(aq) + Ca3(PO4)2 (s)

Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What massof Calcium Phosphate could be formed?

Plan: (1) Calculate moles of P4. (2) Use molar ratios to get moles of Ca3(PO4)2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate.

Calculating the Amounts of Reactants and Products in a Reaction Sequence - II

Solution: moles of P4 =

For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s) 4 P4O10 (s) + 10 KCl (s) ]

For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq) ]

For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2 1 Ca3(PO4)2 + 6 H2O]

moles Ca3(PO4)2 =

moles P4 x

Calculating the Amounts of Reactants and Products in a Reaction Sequence - II

Solution: moles of Phosphorous = 15.50 g P4 x = 0.1251 mol P4

For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s) 4 P4O10 (s) + 10 KCl (s) ]

For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq) ]

For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2 1 Ca3(PO4)2 + 6 H2O]

0.1251 moles P4 x x x

= 0.2502 moles Ca3(PO4)2

1 mole P4

123.88 g P4

4 moles H3PO4

1 mole P4O10

4 moles P4O10

4 moles P4

1 mole Ca3(PO4)2

2 moles H3PO4

Molar mass of Ca3(PO4)2 = 310.18 g mole

Mass of Ca3(PO4)2 product =

Calculating the Amounts of Reactants and Products in a Reaction Sequence - III

Molar mass of Ca3(PO4)2 = 310.18 g mole

mass of product = 0.2502 moles Ca3(PO4)2 x =

= 77.61 g Ca3(PO4)2

Calculating the Amounts of Reactants and Products in a Reaction Sequence - III

310.18 g Ca3(PO4)2

1 mole Ca3(PO4)2

Fig. 3.9 Balanced reaction! Definesstoichiometric ratios!

Unbalanced (i.e., non-stoichiometric) mixture! Limited by syrup!

Limiting Reactant Problems

a A + b B + c C d D + e E + f F

Steps to solve1) Identify it as a limiting reactant problem - Information on the: mass, number of moles, number of molecules, volume and molarity of a solution is given for more than one reactant!2) Calculate moles of each reactant!3) Divide the moles of each reactant by stoic. coefficient (a,b,c etc...)!4) Which ever is smallest, that reactant is the limiting reactant!5) Use the limiting reactant to calculate the moles of product desired then convert to the units needed (moles, mass, volume, number of atoms etc....)!

• 2Al(s) + 6HCl(g) 2AlCl3(s) + 3H2(g)

Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed?

• 30.0 g Al

• 20.0g HCl

• Limiting reactant = one w/ fewer “equivalents” =

Acid - Metal Limiting Reactant - I