Nitrogen forms five compounds with oxygen in which 1.0 g of nitrogen combines with 0.572,1.14,1.73,2.28 and 2.85 g of oxygen respectively. Show that these figures agree with law of multiple proportion.
To show that these figures agree with the law of multiple proportions, we need to determine if the ratios of the masses of oxygen in each compound are in simple whole number ratios.
Let's calculate these ratios for the given compounds:
Compound 1: Nitrogen (N) = 1.0 g, Oxygen (O) = 0.572 g
Ratio of N to O: 1.0 g / 0.572 g ≈ 1.75
Compound 2: Nitrogen (N) = 1.0 g, Oxygen (O) = 1.14 g
Ratio of N to O: 1.0 g / 1.14 g ≈ 0.88
Compound 3: Nitrogen (N) = 1.0 g, Oxygen (O) = 1.73 g
Ratio of N to O: 1.0 g / 1.73 g ≈ 0.58
Compound 4: Nitrogen (N) = 1.0 g, Oxygen (O) = 2.28 g
Ratio of N to O: 1.0 g / 2.28 g ≈ 0.44
Compound 5: Nitrogen (N) = 1.0 g, Oxygen (O) = 2.85 g
Ratio of N to O: 1.0 g / 2.85 g ≈ 0.35
From the calculated ratios, we can see that the values are not in simple whole number ratios. This means that the given data does not strictly follow the law of multiple proportions.
To show that these figures agree with the law of multiple proportions, we need to determine if the ratios of the masses of oxygen present in each compound are in small whole number ratios.
Let's calculate the ratios for each compound:
1. Compound 1: 1.0 g nitrogen and 0.572 g oxygen
Oxygen-to-nitrogen ratio: 0.572 g / 1.0 g = 0.572
2. Compound 2: 1.0 g nitrogen and 1.14 g oxygen
Oxygen-to-nitrogen ratio: 1.14 g / 1.0 g = 1.14
3. Compound 3: 1.0 g nitrogen and 1.73 g oxygen
Oxygen-to-nitrogen ratio: 1.73 g / 1.0 g = 1.73
4. Compound 4: 1.0 g nitrogen and 2.28 g oxygen
Oxygen-to-nitrogen ratio: 2.28 g / 1.0 g = 2.28
5. Compound 5: 1.0 g nitrogen and 2.85 g oxygen
Oxygen-to-nitrogen ratio: 2.85 g / 1.0 g = 2.85
Now, let's examine these ratios to determine if they are in small whole numbers.
Rounding these ratios to two decimal places, we have:
Compound 1: 0.57
Compound 2: 1.14
Compound 3: 1.73
Compound 4: 2.28
Compound 5: 2.85
Based on these values, it may not be immediately clear if the ratios are small whole numbers. However, if we multiply each ratio by a factor that will ensure the smallest whole number ratio, we might be able to see a pattern.
To do this, let's multiply each ratio by 100:
Compound 1: (0.57 * 100) = 57
Compound 2: (1.14 * 100) = 114
Compound 3: (1.73 * 100) = 173
Compound 4: (2.28 * 100) = 228
Compound 5: (2.85 * 100) = 285
Now, we can see that the ratios are approximately 57:114:173:228:285, which simplify to 1:2:3:4:5.
These ratios are indeed in small whole number proportions, confirming that the figures agree with the law of multiple proportions.
A reaction requires pure anhydrous calcium Sulfate CaSO4 only unidentified hydrate of calcium Sulfate CaSO4.xH2O is available
1-we heat 67.59 g of unknown hydrate until all the water has been driven off. The resulting mass of pure CaSO4 is 53.4g what is the formula of the hydrate and what is its formula weight?
2-Suppose we wish to obtain enough of this hydrate to supply 95.5 g of CaSO4. How many grams should we weight out?
Show that the ratios or "proportions" of the number of moles involved in each compound can be expressed as the ratios of small integers.
For example:
1.0/14 = 0.07143 (moles of N)
0.572/16 = 0.03575 (moles of O)
The ratio N/O is 1.998 or (rounding off to 3 figures), 2:1. it corresponds to nitrous oxide, N2O.
The others should round off to molar N:O proportions of
1:1, 1:2, 2:3 and 2:5.