Determine the length of the object shown. 7 cm 7.2 cm 72.5 mm 72.50 mm
2 answers:
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Answer:
Question 7 to 12 are given in attached file because character limit is only 5000
Explanation:
1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of sulfur. Calculate the empirical formula of this compound.
Given data:
Mass of sample = 15 g
Mass of sodium = 8.83 g
Mass of sulfur = 6.17 g
Empirical formula = ?
Solution:
Number of gram atoms of Na = 8.83 / 23 = 0.4
Number of gram atoms of S = 6.17 / 32 = 0.2
Atomic ratio:
Na : S
0.4/0.2 : 0.2/0.2
2 : 1
Na : S = 2 : 1
Empirical formula is Na₂S.
2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?
Given data:
Mass of phosphorus = 4.433 g
Mass of oxygen = 10.150 g - 4.433 g = 5.717 g
Empirical formula = ?
Solution:
Number of gram atoms of P = 4.433 / 30.9738 = 0.1431
Number of gram atoms of O = 5.717/ 15.999 = 0.3573
Atomic ratio:
P : O
0.1431/0.1431 : 0.3573/0.1431
1 : 2.5
P : O = 2(1 : 2.5)
Empirical formula is P₂O₅.
3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.
Given data:
Percentage of sodium = 36.48%
Percentage of sulfur = 25.41%
Percentage of oxygen = 38.11%
Empirical formula = ?
Solution:
Number of gram atoms of Na = 36.48 / 23 = 1.6
Number of gram atoms of S = 25.41/ 32 = 0.8
Number of gram atoms of O = 38.11/ 16 = 2.4
Atomic ratio:
Na : S : O
1.6/0.8 : 0.8/0.8 : 2.4/0.8
2 : 1 : 3
Na: S : O = 2 : 1 : 3
Empirical formula is Na₂SO₃.
4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.
Given data:
Percentage of iron = 63.52%
Percentage of sulfur = 36.48%
Empirical formula = ?
Solution:
Number of gram atoms of Fe = 63.52 / 55.845 = 1.14
Number of gram atoms of S = 36.48 / 32 = 1.14
Atomic ratio:
Fe : S
1.14/1.14 : 1.14/1.14
1 : 1
Fe : S = 1 : 1
Empirical formula is FeS.
5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 % oxygen. Find the empirical formula of this compound.
Given data:
Percentage of sodium = 32.38%
Percentage of sulfur = 22.65%
Percentage of oxygen = 44.99%
Empirical formula = ?
Solution:
Number of gram atoms of Na = 32.38 / 23 = 1.4
Number of gram atoms of S = 22.65/ 32 = 0.7
Number of gram atoms of O = 44.99/ 16 = 2.8
Atomic ratio:
Na : S : O
1.4/0.7 : 0.7/0.7 : 2.8/0.7
2 : 1 : 4
Na: S : O = 2 : 1 : 4
Empirical formula is Na₂SO₄.
6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that 4.00 grams of calcium are present. What is the empirical formula of the compound formed?
Given data:
Mass of sample = 20g
Mass of bromine = 20 g - 4 g = 16 g
Mass of calcium = 4 g
Empirical formula = ?
Solution:
Number of gram atoms of bromine = 16 / 80= 0.2
Number of gram atoms of calcium = 4/ 40= 0.1
Atomic ratio:
Ca : Br
0.1/0.1 : 0.2/0.1
1 : 2
Ca: Br = 1 : 2
Empirical formula is CaBr₂.
