Percent Composition and Chemical Formulas

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Transcript Percent Composition and Chemical Formulas

PERCENT COMPOSITION AND CHEMICAL FORMULAS

03/09 – 03/12

Percent Composition by Mass

π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘’π‘™π‘’π‘šπ‘’π‘›π‘‘ % 𝑏𝑦 π‘šπ‘Žπ‘ π‘  = π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘π‘œπ‘šπ‘π‘œπ‘’π‘›π‘‘ Γ— 100

Ex: Percent Composition by Mass

 Given: 100 g sample 55 g Element X 45 g Element Y  Work: X: % π‘π‘œπ‘šπ‘ πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ 𝑋 = 55 𝑔 πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ 𝑋 Γ— 100 100 𝑔 % π‘π‘œπ‘šπ‘ πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ 𝑋 = 55% πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ 𝑋 Y: % π‘π‘œπ‘šπ‘ πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ π‘Œ = 45 𝑔 πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ π‘Œ Γ— 100 100 𝑔 % π‘π‘œπ‘šπ‘ πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ π‘Œ = 45% πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘ π‘Œ

Example Cont

 Check your work: The sum of the percentages should add up to 100: 55% Element X + 45% Element Y = 100%

Pg 350, problem 62

49.98 g O 10.47 g H 60.45 g sample What is the percent composition by mass of each element?

Pg 350, problem 63

46.45% N 53.32% O 100 g sample What are the masses of N and O?

Pg 350, problem 65

65.4% C 29.09% O 5.45% H 23.8 g sample What are the masses of each element?

Percent Composition from Chemical Formula % π‘π‘œπ‘šπ‘ 𝑋 = π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘šπ‘Žπ‘ π‘  π‘œπ‘“ 𝑋 π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘π‘œπ‘šπ‘π‘œπ‘’π‘›π‘‘ Γ— 100

Example: pg 343 in Text book

  1.

2.

Given: NaHCO 3 What is the percent composition of each element?

Find the molar mass of each element: 1.

2.

3.

4.

Na: 22.98 g/mole H: 1.01 g/mole C: 12.01 g/mole O: 15.99 g/mole 1.

Find the molar mass of the compound: 22.98 g/mole + 1.01 g/mole + 12.01 g/mole + 3(15.99 g/mole) = 83.97 g/mole 3.

4.

Find the percent composition for each element: 1.

Na: 22.98 𝑔/π‘šπ‘œπ‘™π‘’ 83.97 𝑔/π‘šπ‘œπ‘™π‘’ 27.37% Γ— 100 = 2.

3.

4.

H: 1.01 𝑔/π‘šπ‘œπ‘™π‘’ 83.97 𝑔/π‘šπ‘œπ‘™π‘’ Γ— 100 = 1.20% C: 12.01 𝑔/π‘šπ‘œπ‘™π‘’ 83.97 𝑔/π‘šπ‘œπ‘™π‘’ O: 3(15.999

𝑔 π‘šπ‘œπ‘™π‘’ ) 83.97 𝑔/π‘šπ‘œπ‘™π‘’ Γ— 100 = 14.30% Γ— 100 = 57.13% Check your work: the sum of the percentages should be between 99 and 101: 27.37% + 1.20% + 14.30% + 57.13% = 100%

Pg 344 #54

What is the percent composition of each element in H 3 PO 4 ?

Pg 344, #56

Calculate the percent by mass of each element in CaCl 2.

Empirical Formulas

 Formula for a compound with the smallest whole number mole ratio of elements  Method for determining the subscripts in a formula

Ex: Determining the Empirical Formula Given Percent Composition You have a 100 g sample made up of S and O. 40.05% of this compound is S, while 59.95% is O. What is the empirical formula?

1.

Figure out how many grams of each element you have: S: 40.05% 𝑆 = 𝑋 𝑔 𝑆 Γ— 100 100𝑔 40.05 𝑔 = 𝑋 𝑔 𝑆 O: 59.95% 𝑂 = 𝑋 𝑔 𝑂 100 𝑔 Γ— 100 59.95 𝑔 = 𝑋 𝑔 𝑂

Clicker Question 1

If you have a 20 g sample that is 43% Oxygen, how many grams of Oxygen do you have?

A. 8.75 g B. 86 g C. 215 g D. Not enough information

Clicker Question 2

A 15 g sample contains 8.32 g of S. What is its percent composition of S?

A. 55.47% B. 62.71% C. 58.95% D. Not enough information

Clicker Question 3

A 48 g sample is made up of C, O, and H. Its percent composition of C is 23%. What is the mass of O in the compound?

A. 11.04 g B. 47.92 g C. 38.72 g D. Not enough information

Ex: Determining the Empirical Formula Given Percent Composition cont.

2. Using the molar mass of each element, determine how many moles of each element you have: S: 40.05 𝑔 𝑆 Γ— 1 π‘šπ‘œπ‘™π‘’ 𝑆 32.07 𝑔 𝑆 O: 59.95 𝑔 𝑂 Γ— 1 π‘šπ‘œπ‘™π‘’ 𝑂 16 𝑔 𝑂 = 1.249 π‘šπ‘œπ‘™π‘’ 𝑆 = 3.747 π‘šπ‘œπ‘™π‘’ 𝑂

Clicker Question 4

Which letter represents the molar mass of this element?

A B C D

Clicker Question 5

What are the appropriate units of molar mass?

A. g B. Moles C. g/mole D. mole/g

Ex: Determining the Empirical Formula Given Percent Composition cont.

3. Figure out which element has the smallest number of moles: S, with 1.249 moles 4. Divide the moles of each element by the smallest number of moles: S: 1.249 π‘šπ‘œπ‘™π‘’ 𝑆 1.249 π‘šπ‘œπ‘™π‘’ O: 3.747 π‘šπ‘œπ‘™π‘’ 𝑂 1.249 π‘šπ‘œπ‘™π‘’ = 1𝑆 = 3𝑂 5. These numbers are now the subscripts for your empirical formula: SO 3 *If dividing by the smallest mole value does not give you a whole #, then you must find the least common multiple of the mole values*

Clicker Question 6

In the previous example, we determined that the empirical formula for the compound was SO 3 . How many Oxygen atoms are there in one molecule?

A.

1 B.

2 C.

D.

3 4

1.

2.

3.

4.

5.

Summary of Finding Empirical Formula Steps Figure out how many grams of each element you have Using the molar mass of each element, determine how many moles of each element you have Figure out which element has the smallest number of moles Divide the moles of each element by the smallest mole value (or multiply if dividing does not give you whole numbers) Use the results of that division to assign subscripts.

Finding Empirical Formula: Example to Work Through as a Class 100 g sample made up of C, H, and O 48.64% C 8.16% H 43.20% O What is the empirical formula for this compound?

Pg 350, # 64

23.55 g of a compound is decomposed, resulting in 83.01% K and 16.99 O. What is the empirical formula for this compound?

Pg 350, # 66

A 100 g sample of morphine is broken down and yields the data in the table. What is the empirical formula for morphine?

Element

C H O N

Mass (g)

17.900

1.680

4.225

1.228