Afra Khanani Period 6 Honors Chemistry March 31 st
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Afra Khanani Period 6 Honors Chemistry March 31 st
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Afra Khanani Period 6 Honors Chemistry March 31 st. PROBLEM:. A solution containing 6720 mg of H 2 0 is added to another containing 10.67 Liters of CO 2 at STP. - PowerPoint PPT Presentation
Transcript of Afra Khanani Period 6 Honors Chemistry March 31 st
Afra KhananiPeriod 6Honors ChemistryMarch 31st
PROBLEM:A solution containing 6720 mg of H20 is added to
another containing 10.67 Liters of CO2 at STP.
Determine which reactant was in excess, as well as the number of grams over the amount required by
the limiting species. Also, find the number of molecules of glucose that precipitated as well, as
the theoretical and percent yield of glucose if 10.22 g C6H12O6 was obtained.
The chemical reaction of water and carbon dioxide will produce oxygen gas, and an
unknown element. This unknown element has an empirical
formula of CH2O and a molecular mass of 180.15. Its molecular formula is the unknown
product needed for your equation.
Products
STEP 1Find the unknown product
STEP 1Find the unknown product
Givens: Empirical formula: CH2O Molecular mass: 180.15
STEP 1Find the unknown product
Givens: Empirical formula: CH2O Molecular mass: 180.15
1. Find the mass of your empirical formula: CH2O = 30g
STEP 1Find the unknown product
Givens: Empirical formula: CH2O Molecular mass: 180.15
1. Find the mass of your empirical formula: CH2O = 30g
2. Divide the molecular mass by the empirical mass: 180/30 = 6
STEP 1Find the unknown product
Givens: Empirical formula: CH2O Molecular mass: 180.15
1. Find the mass of your empirical formula: CH2O = 30g
2. Divide the molecular mass by the empirical mass: 180/30 = 6
3. Multiply that number (6) to your empirical formula (CH2O) : C6H12O6
STEP 1Find the unknown product
Givens: Empirical formula: CH2O Molecular mass: 180.15
1. Find the mass of your empirical formula: CH2O = 30g
2. Divide the molecular mass by the empirical mass: 180/30 = 6
3. Multiply that number (6) to your empirical formula (CH2O) : C6H12O6
Molecular Formula and Unknown Product= C6H12O6
STEP 1Write and balance the equation
STEP 1Write and balance the equation
__ H20 + __ CO2 __ C6H12O6 + __ 02
REACTANTS PRODUCTS
STEP 1Write and balance the equation
__ H20 + __ CO2 __ C6H12O6 + __ 02
REACTANTS PRODUCTS
Water + Carbon Dioxide (+ energy) = Glucose + Oxygen
STEP 2Start with one of the knowns (convert mg to g)
6720 mg of H20 (Given)
STEP 2Start with one of the knowns (convert mg to g)
6720 mg of H20 (Given)
6720 mg H20 1 gram H20
1000 milligrams H20
1 gram H20 = 1000 mg H20
STEP 3Convert grams to moles
STEP 3Convert grams to moles
6.72 g H20 1 mole H20
18 grams H20
1 mole H20 = 18 g H20
STEP 4Convert mole to moles
6 H20 + 6 CO2 C6H12O6 + 6 02
STEP 4Convert mole to moles
STEP 4Convert mole to moles
6 H20 + 6 CO2 C6H12O6 + 6 02
.373 mole H20 1 mole C6H12O6
6 mole H20
6 mole H20 = 1 mole C6H1206
STEP 5Convert moles to grams
.062 mole C6H1206 180 grams C6H1206
1 mole C6H1206
STEP 5Convert moles to grams
1 mole C6H1206 = 180 g C6H1206
You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6
Now figure out: 10.67 L CO2 = ? grams C6H12O6
STEP 1Convert L at STP to moles
10.67 L of CO2 (Given)
STEP 1Convert L at STP to moles
10.67 L of CO2 (Given)
10.67 L CO2 1 mole CO2
22.4 Liters CO2
22.4 L CO2 =1 mole CO2
STEP 2Convert moles to moles
6 H20 + 6 CO2 C6H12O6 + 6 02
STEP 2Convert moles to moles
6 H20 + 6 CO2 C6H12O6 + 6 02
.476 mole CO2 1 mole C6H12O6
6 mole CO2
6 mole CO2 = 1 mole C6H1206
STEP 3Convert moles to grams
STEP 3Convert moles to grams
.079 mole C6H12O6 180 grams C6H12O6
1 mole C6H12O6
1 mole C6H1206 = 180 g C6H1206
You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6
10.67 L of CO2 = 14.22 mol C6H12O6
CO2 is the excess reactant
You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6
10.67 L of CO2 = 14.22 mol C6H12O6
CO2 is the excess reactant
How much excess CO2 ?(In grams)
CO2 is the excess reactant
How much excess CO2 ?(In grams)
14.22 grams – 11.16 grams =
You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6
10.67 L of CO2 = 14.22 mol C6H12O6
3.06 grams CO2 in excess
Find the number of molecules of glucose that precipitated.
What’s Next?
STEP 1Convert moles to molecules
0.62 mole of C6H12O6 (Found)
STEP 1Convert moles to molecules
0.62 mole of C6H12O6 (Found)
0.62 moles C6H12O6 6.02 x 1023 C6H12O6
1 mole C6H12O6
1 mole C6H1206 = 6.02 x 1023 molecules C6H1206
Find the number of molecules of glucose that precipitated.
RESTATING THE QUESTION:
Find the number of molecules of glucose that precipitated.
3.73 E22 molecules C6H12O6
RESTATING THE QUESTION:
THEORETICAL & PERCENT YIELDFind theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
THEORETICAL & PERCENT YIELD
Theoretical Yield: (Already Found) 11.16 grams
Find theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
THEORETICAL & PERCENT YIELD
Theoretical Yield: (Already Found) 11.16 grams
Percent Yield: ACTUAL YIELD/THEORETICAL x 100
Find theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
THEORETICAL & PERCENT YIELDFind theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
Theoretical Yield: (Already Found) 11.16 grams
Percent Yield: ACTUAL YIELD/THEORETICAL x 100
PERCENT YIELD: 10.22/11.16 x 100 = 92%
Percent Error%Error =
(|Your Result - Accepted Value| / Accepted Value) x 100
Percent Error%Error =
(|Your Result - Accepted Value| / Accepted Value) x 100
How much should have been made = 11.16 g GlucoseHow much was made: 10.22 g Glucose
Percent Error
%Error =(|Your Result - Accepted Value| / Accepted Value) x 100
How much should have been made = 11.16 g GlucoseHow much was made: 10.22 g Glucose
(|10.22 – 11.16| / 11.16) x 100 = 8%
