What is the relative formula mass of: NaBr, CO2, H2S, BaSO4, Cu(NO3)2, FeCL3?
The relative formula mass is the sum of the Ar of all of the elements in the compound. The Ar can be found on the periodic table, as is the larger of the two numbers associated with the element. For NaBr, this would be 1xNa = 22.989 + 1xBr = 79.9... This leaves the Mr as 102.889. For CO2 there are 2 x Oxygen, so the sum becomes (2x16) + 12 = Mr of 44. The same for H2S, as there are 2xH molecules, (2x1) + 32 = 34. For BaSO4 there are 4 Oxygens and one of everything else, (4x16)+ 32+ 137.327 = 233.327. For FeCl3 there are 3 chlorines and one Fe, (3x 35.4) + 55.8 = 162 For Cu(NO3)2, this becomes more complicated. Everything within the brackets is times by 2. So we are left with 1X Cu, 2xN, and for O (which is already times by 3 within the bracket and then times by 2 from outside) 6xO. Therefore in that order... 63.5 + (14x2) + (6x16) = 187.5. I hope this was helpful :) obviously the exact answer given with depend on the amount of significant figures or decimal points are requested in the question!
Relative formula mass Relative atomic masses can be used to find the relative formula mass of a compound. To find the relative formula mass (Mr) of a compound, you add together the relative atomic mass values (Ar values) for all the atoms in its formula. Here is one example: Find the Mr of carbon monoxide, CO. The Ar of carbon is 12 and the Ar of oxygen is 16. The Mr of carbon monoxide is 12 + 16 = 28. Example 2: Find the Mr of Copper Nitrate Cu(NO3)2 Ar of Cu = 63.5. Ar of N = 14 x 2. Ar of O = 6 x 16. Remember that the nitrate ion contains 3 oxygen atoms. But there are are 2 nitrate ions in copper nitrate, so there are 2 lots of 3 oxygen atoms. THAT IS WHY YOU MULTIPLY 6 BY 16. There is only one nitrogen atom in a nitrate ion, but there are 2 nitrate ions in copper nitrate. SO YOU MULTIPLY 2 X 14. Add up all of the relative atomic masses = 63.5+(14x2) + (16x6) = 187.5
NaBr: 23 + 79.9 = 102.9 CO2: 12 + 16 + 16 = 44 H2S: 1 + 1 + 32.1 = 34.1 BaSO4: 137.3 + 32.1 + 16 + 16 + 16 + 16 = 233.4 Cu(NO3)2: 63.5 + 14 + 16 + 16 + 16 + 14 + 16 + 16 + 16 = 187.5 FeCl3: 55.8 + 35.5 + 35.5 + 35.5 = 162.3
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