Carbon calculations!!!

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# Carbon calculations!!! - PowerPoint PPT Presentation

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

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### Carbon calculations!!!

What’s your usage/day in grams CO2?

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

end so it’s C9H13O3N)

Combustion Train for the Determination of the Chemical Composition of Organic Compounds.

m

2

m

2

CnHm + (n+ ) O2 = n CO(g) + H2O(g)

Fig. 3.4

Ascorbic Acid ( Vitamin C ) - I Contains 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)

= 2.92 x 10-4 g H

• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg

= 3.54 mg 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 2.21 x 10-4
• C = 1.00 Multiply each by 3 = 3.00 = 3.0
• H = 1.32 = 3.96 = 4.0
• O = 1.00= 3.00 = 3.0

C3H4O3

Determining a Chemical Formula from

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

molecular formula.

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 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

Determining a Chemical Formula from

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

Some Compounds with Empirical Formula

CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O)

Molecular M

Formula (g/mol) Name Use or Function

CH2O 30.03 Formaldehyde Disinfectant; Biological

preservative

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

metabolism

C5H10O5 150.13 Ribose Component of many nucleic

acids and vitamin B2

C6H12O6 180.16 Glucose Major nutrient for energy

in cells

Two Compounds with Molecular Formula C2H6O

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

alcoholic beverages

H H H H

H C C O H H C O C H

H H H H

Table 3.4

Molecular

Formula

Atoms

Molecules

Number

6.022 x 1023

Moles

Moles

Chemical Equations

Qualitative Information:

Reactants

Products

States of Matter: (s) solid

(l) liquid

(g) gaseous

(aq) aqueous

2 H2 (g) + O2 (g) 2 H2O (g)

Chemical Equation Calculation - I

Atoms (Molecules)

Number

6.02 x 1023

Molecules

Reactants

Products

Chemical Equation Calculation - II

Mass

Atoms (Molecules)

Molecular

Weight

Number

g/mol

6.02 x 1023

Molecules

Reactants

Products

Moles

Stoichiometry

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.

Information Contained in a Balanced Equation

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

1. Convert grams of A to moles of A

 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

• To find the amount of B (one reactant or product) given the amount of A (another reactant or product):
Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

How many grams of CO2 are produced when 20.0 g of propane is burned?

Molar masses

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)

Limiting Reactant

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.

All amounts produced and reacted are determined by the limiting reactant.

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.

Magnesium metal is used to prepare zirconium metal, which is used to make the container for nuclear fuel (the nuclear fuel rods):

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?

ZrCl4 is the limiting reactant.

0.20 mol Zr will be produced.

Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature:

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?

Molar masses

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.

To find the excess NH3, we find how much NH3 reacted:

Now subtract the amount reacted from the starting amount:

10.0 at start

-7.73 reacted

2.27 g remains

2.3 g NH3 is left unreacted.

(1 decimal place)

Theoretical Yield

The maximum amount of product that can be obtained by a reaction from given amounts of reactants. This is a calculated amount.

Actual Yield

The amount of product that is actually obtained. This is a measured amount.

Percentage Yield

2NH3 + CO2(g)  CH4N2O + H2O

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)