Formula mass for class 10 ok1294987900
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![Relative formula mass, M r The relative formula mass of a compound is blatantly the relative atomic masses of all the elements in the compound added together. E.g. water H 2 O: Therefore M r for water = 16 + (2x1) = 18 Work out M r for the following compounds: ,[object Object],[object Object],[object Object],[object Object],[object Object],H=1, Cl=35 so M r = 36 Na=23, O=16, H=1 so M r = 40 Mg=24, Cl=35 so M r = 24+(2x35) = 94 H=1, S=32, O=16 so M r = (2x1)+32+(4x16) = 98 K=39, C=12, O=16 so M r = (2x39)+12+(3x16) = 138 Relative atomic mass of O = 16 Relative atomic mass of H = 1](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Formula mass for class 10 ok1294987900
- 1. Relative mass formula, atomic mass, and empirical formula
- 3. More examples CaCO 3 40 + 12 + 3x16 100 HNO 3 1 + 14 + 3x16 2MgO 2 x (24 + 16) 80 3H 2 O 3 x ((2x1) + 16) 4NH 3 2KMnO 4 3C 2 H 5 OH 4Ca(OH) 2
- 8. Another method Try using this equation: Mass of product IN GRAMMES 6g 4 36 So mass of product = (4/36) x 6g = 0.66g of hydrogen
- 11. Empirical formulae Empirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO 3 , H 2 0 and KMnO 4 are all empirical formulae. Here’s how to work them out: A classic exam question: Find the simplest formula of 2.24g of iron reacting with 0.96g of oxygen. Step 1: Divide both masses by the relative atomic mass: For iron 2.24/56 = 0.04 For oxygen 0.96/16 = 0.06 Step 2: Write this as a ratio and simplify: 0.04:0.06 is equivalent to 2:3 Step 3: Write the formula: 2 iron atoms for 3 oxygen atoms means the formula is Fe 2 O 3