Unit 1: 2.2: Laws of Chemical Combination Flashcards
The composition of a compound is fixed; also called the law of constant composition.
the law of definite proportions
The ratio of the number of a certain item divided by the total number of items.
proportion
The amount of a certain component in 100 units of the total.
percent
When two or more elements combine to from more than one compound, for a fixed mass of one element, the masses of each of the other elements in the compounds occurs in a small, whole-number ratio.
the law of multiple properties
During a chemical reaction, mass is neither gained nor lost.
the law of conservation of mass
A 4.33 g sample of dinitrogen monoxide, N2O, is composed of 63.65% nitrogen and 36.35% oxygen by mass. What is the percent composition of a 14.9 g sample of N2O?
the 14.9g sample is 63.65% nitrogen and 36.35% oxygen by mass also. According to the law of definite proportions, all pure samples of N2O have the same percent composition.
Nitrogen monoxide, NO, has a percent composition of 46.68% nitrogen and 53.32% oxygen by mass. What possible percentages of nitrogen could be in a mixture of N2O and NO?
The possible percentages of nitrogen in the N2O/NO mixture would be between 63.65% (the percentage of nitrogen in N2O) and 46.68 (the percentage of nitrogen in NO).
A sample of 25.50g of dinitrogen monoxide, N2O, decomposes into 9.866g of nitrogen and 5.634g of oxygen. A sample of 25.50g N2O decomposes into 16.23g of nitrogen and 9.269g of oxygen. Show that the amount of nitrogen and oxygen obtained in each decomposition is consistent with the law of definite proportions.
For the 15.50g sample:
(9.866g N/ 15.50 g N2O) x 100%= 63.65% N
(5.634g O/ 15.59g N2O) x 100%= 36.35% O
For the 25.50g sample:
(16.23g N/ 25.50g N2O) x 100%= 63.65%N
(9.269g O/ 25.50g N2O) x 100%= 36.35%O
The percentages are the same for both samples, which is consistent with the law of definite proportions. Percent composition by mass is independent of the total mass of the sample.
Calculate the mass of nitrogen in 4.75g sample of nitrogen monoxide, NO, using the percentages given:
N2O consists of 63.65% nitrogen by mass and 36.65% oxygen by mass.
mass of element= (% by mass/ 100%) x mass of sample
mass of N = (46.68%/100%)(4.75g)= 2.22g
Given that NO is 46.68% nitrogen by mass, calculate the mass of NO that contains 100.0g of nitrogen
100.0g N (100.0g NO/ 46.68g N) = 214.2g NO
A sample of nitrogen monoxide, NO, consists of 14.01g of nitrogen and 16.00g of oxygen, whereas a sample of nitrogen dioxide, NO2, consists of 14.01g of nitrogen and 32.00g of oxygen. Show that NO and NO2 follow the law of multiple proportions.
Calculate the nitrogen to oxygen (N:O) ratio for both compounds:
for NO: (14.01g N/ 16.00g O)= 0.8756
for NO2: (14.01g N/32.00g O)= 0.4378
the N:O mass ratios of these two compounds can now be compared. It is easiest to divide the larger value by the smaller value:
(N:O ratio for NO/N:O ratio for NO2) = (0.8756/0.4378) = 2.000
The ratio of N:O mass ratios for these compounds is a whole number, 2, so NO and NO2 follow the law of multiple proportions.
A sample of sulfur dioxide, SO2, consists of 16.04g of sulfur and 16.00g of oxygen, whereas a sample of sulfur trioxide, SO3, consists of 16.04g of sulfur and 24.00g of oxygen. Show that SO2 and SO3 follow the law of multiple proportions.
Calculate the sulfur-to-oxygen, S:O, ratios for both compounds:
for SO2: (16.04g S/ 16.00g O) = 1.003
for SO3: (16.04g S/ 24.00g O)= 0.6683
The S:O mass ratios of these two compounds can now be compared. Dividing the larger value by the smaller value gives
(S:O ratio for SO2/ S:O ratio for SO3) = (1.003/0.6683) = 1.501
multiplying 1.501 by 2 gives a ratio of 3.002
The ratio of S:O mass ratios for these compounds is a whole number, 3, so SO2 and SO3 follow the law of multiple proportions.
