Carbon calculations!!!. What’s your usage/day in grams CO 2 ?. Me: . You're pretty low impact - nice work! Water 19,200 GAL Energy 115 MBTU Carbon Dioxide 22,000 LBS Wastewater 19,000 GAL Runoff 121,300 GAL Trash 330 LBS Money 2,890 $. (This has an extra methyl group on the
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
What’s your usage/day in grams CO2?
You're pretty low impact - nice work!
methyl group on the
end so it’s C9H13O3N)
Combustion Train for the Determination of the Chemical Composition of Organic Compounds.
CnHm + (n+ ) O2 = n CO(g) + H2O(g)
= 2.65 x 10-3 g C
= 2.92 x 10-4 g H
= 3.54 mg O
= 2.21 x 10-4 mol C
= 2.92 x 10-4 mol H
= 2.21 x 10-4 mol O
Combustion Analysis - I
Problem: Erthrose (M = 120 g/mol) is an important chemical
compound 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
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 xM of C
mass of 1 mol CO2
1 mol C x 12.01 g C/ 1 mol C
44.01 g CO2
mol H xM of H
mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H
18.02 g H2O
Combustion Analysis - III
0.2729 g C
1 g CO2
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 O
Calculating 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 / formula
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
0.1119 g H
1 g H2O
CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O)
Formula (g/mol) Name Use or Function
CH2O 30.03 Formaldehyde Disinfectant; Biological
C2H4O2 60.05 Acetic acid Acetate polymers; vinegar
( 5% solution)
C3H6O3 90.08 Lactic acid Causes milk to sour; forms
in muscle during exercise
C4H8O4 120.10 Erythrose Forms during sugar
C5H10O5 150.13 Ribose Component of many nucleic
acids and vitamin B2
C6H12O6 180.16 Glucose Major nutrient for energy
Property Ethanol Dimethyl Ether
M (g/mol) 46.07 46.07
Color Colorless Colorless
Melting point - 117oC - 138.5oC
Boiling point 78.5oC - 25oC
Density (at 20oC) 0.789 g/mL 0.00195 g/mL
Use Intoxicant in In refrigeration
H H H H
H C C O H H C O C H
H H H H
6.022 x 1023
States of Matter: (s) solid
2 H2 (g) + O2 (g) 2 H2O (g)
6.02 x 1023
6.02 x 1023
The calculation of the quantities of reactants and products involved in a chemical reaction
Interpreting a Chemical Equation
The coefficients of the balanced chemical equation may be interpreted in terms of either (1) numbers of molecules (or ions or formula units) or (2) numbers of moles, depending on your needs.
Viewed in Reactants Products
terms 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 H2O
Amount (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
Using the molar mass of A
2. Convert moles of A to moles of B
Using the coefficients of the balanced chemical equation
3. Convert moles of B to grams of B
Using the molar mass of B
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
How many grams of CO2 are produced when 20.0 g of propane is burned?
C3H8: 3(12.01) + 8(1.008) = 44.094 g
CO2: 1(12.01) + 2(16.00) = 44.01 g
59.9 g CO2
(3 significant figures)
The reactant that is entirely consumed when a reaction goes to completion
Once one reactant has been completely consumed, the reaction stops.
Any problem giving the starting amount for more than one reactant is a limiting reactant problem.
How can we determine the limiting reactant?
Use each given amount to calculate the amount of product produced.
The limiting reactant will produce the lesser or least amount of product.
ZrCl4(g) + 2Mg(s) 2MgCl2(s) + Zr(s)
How many moles of zirconium metal can be produced from a reaction mixture containing 0.20 mol ZrCl4 and 0.50 mol Mg?
0.20 mol Zr will be produced.
2NH3 + CO2(g) CH4N2O + H2O
In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions?
NH3 1(14.01) + 3(1.008) = 17.02 g
CO2 1(12.01) + 2(16.00) = 44.01 g
CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g
CO2 is the limiting reactant.
13.6 g CH4N2O will be produced.
Now subtract the amount reacted from the starting amount:
10.0 at start
2.27 g remains
2.3 g NH3 is left unreacted.
(1 decimal place)
The maximum amount of product that can be obtained by a reaction from given amounts of reactants. This is a calculated amount.
The amount of product that is actually obtained. This is a measured amount.
When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield?
Theoretical yield = 13.6 g
Actual yield = 9.3 g
= 68% yield
(2 significant figures)