First you want to think "if I had 100 grams of this compound". If you do this, then you will be able to use the percentages as measurements in grams. Completing the sentence: "if I had 100 grams of this compound, then I would have 26 grams of nitrogen and 74 grams of oxygen". The next step is converting the grams to moles.
Grams ÷ Atomic weight = moles
26.0 grams ÷ 14.0 grams = 1.86 moles N
74.0 grams ÷ 16.0 grams = 4.63 moles O
This is a ratio of 1.86:4.63 and is not recognizable as a small whole number ratio. (If this doesn't make sense, see the Law of Definite Proportions). In order to make the two numbers whole numbers, a trick is to divide both by the smallest number. Doing this will ensure that one number will always be 1.
1.86 moles N ÷ 1.86 = 1
4.63 moles O ÷ 1.86 = 2.49
Because the last digit is always uncertain to .02, we can say that 2.49 is actually 2.5. Now your ratio is 1:2.5, yet these are still not whole numbers. When you have an increment of .5, then multiply everything by 2.
1:2.5 × 2 = 2:5 or 2 nitrogen : 5 oxygen
Your final answer is N2O5
The formula of the compound and the Atomic Mass of its elements.
The formula shows that there are three times as many atoms of hydrogen as of nitrogen in the compound. The gram atomic masses are 1.00794 for hydrogen and 14.0067 for nitrogen. Therefore, the percent by mass of hydrogen in the compound is 100{[3(1.00794)]/[3(1.00794) + 14.0067] or 17.7553 %, to the justified number of significant digits.
Fe2S3, I think...
CaBr2 :)
CaS
it can be calculated using the formula percentage composition of N =Gram molecular mass of nitrogen in the compound/ Gram molecular mass of compound *100
The oxide N2O3 has a lower percent of mass nitrogen.
The gram Atomic Mass of cesium is 132.905 and that of nitrogen is 14.0067. The formula of the compound shows that there are three atoms of cesium for each atom of nitrogen. Therefore, the percent of cesium in the compound is: 100{3(132.905)/[3(132.905) + 14.0067]} or 96.6063 % cesium, to the justified number of significant digits. By difference the per cent nitrogen is 3.3937.
The gram atomic mass of cesium is 132.905 and that of nitrogen is 14.0067. The formula of the compound shows that there are three atoms of cesium for each atom of nitrogen. Therefore, the percent of cesium in the compound is: 100{3(132.905)/[3(132.905) + 14.0067]} or 96.6063 % cesium, to the justified number of significant digits. By difference the per cent nitrogen is 3.3937.
The atomic mass of hydrogen is 1 and that of carbon is 12, to the nearest integer. Therefore, a compound that contains 20 percent hydrogen and 80 percent carbon contains atoms of carbon in the ratio of 20:(80/12) = 20:(20/3) = 20 X 3/20 = 3. The smallest whole number ratio corresponding to this is 1:3, which corresponds to an empirical formula of CH3. (However, since carbon normally has a valence of 4 in hydrocarbons, the actual formula is probably C2H6.)
The formula of the compound and the Atomic Mass of its elements.
P2o5
Chi a+
p2o5
S2o3
KCl
Cr3Si2 is the empirical formula for a compound containing chromium and silicon an has 73.52 mass percent chromium.