1 / 7

STANDARD GRADE CHEMISTRY CALCULATIONS

mass n gfm. You must learn this and be able to apply it in calculations. STANDARD GRADE CHEMISTRY CALCULATIONS. Calculations involving the mole. 1 mole of a solid substance is the formula mass of the substance in grams. This is known as the gram formula mass (gfm).

jchristman
Download Presentation

STANDARD GRADE CHEMISTRY CALCULATIONS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. mass n gfm You must learn this and be able to apply it in calculations. STANDARD GRADE CHEMISTRYCALCULATIONS Calculations involving the mole. 1 mole of a solid substance is the formula mass of the substance in grams. This is known as the gram formula mass (gfm). The triangle shown below can be used to give the following relationships:- mass = mass of substance. n = number of moles gfm = gram formula mass • Mass = number of moles x gram formula mass. • 2. Number of moles = mass gram formula mass

  2. mass n gfm Worked example 1. Calculate the mass of 0.25 moles of butane (C4H10). Step 1 :- Write the formula for butane C4H10 Step 2:- Calculate the gram formula mass (4 x 12) + (10 x 1) = 58g Step 3:- Using the triangle we have mass = number of moles x gram formula mass mass = 0.25 x 58 mass = 14.5 g Calculations for you to try. 1. Calculate the mass present in 2.5 moles of calcium carbonate (CaCO3). 2. Calculate the mass of ammonium sulphate, (NH4) 2SO4, present in 0.1 mol of ammonium sulphate? gfm = 100g mass = 2.5 x 100 = 250g gfm = 132g mass = 0.1 x 132 = 13.2g Standard Grade Chemistry

  3. Step 3:- Using the triangle we have number of moles = = = 0.05 mole mass gram formula mass 5.05 101 mass n gfm Worked example 2. Calculate the number of moles in 5.05 g of potassium nitrate, (KNO3). Step 1 :- Write the formula for potassium nitrate KNO3 Step 2:- Calculate the gram formula mass (1 x 39) + (1 x 14) + (3 x 16) = 101g Calculations for you to try. 1. Calculate the number of moles in 132 g of carbon dioxide, CO2. 2. Calculate the number of moles in 4g of bromine, Br2. gfm = 44g Number of moles = 132/44 = 3 moles gfm = 160 Number of moles = 4/160 = 0.025 moles Standard Grade Chemistry

  4. Remember this is in litres n C V (l) You must learn this and be able to apply it in calculations. Mole calculations involving solutions. The concentration of a solution is measured in moles per litre (mol/l) The triangle shown below can be used to give the following relationships:- n = number of moles. C = concentration. V(l) = volume in litres 1. number of moles = concentration x volume (in litres). 2. concentration = number of moles volume (in litres) Standard Grade Chemistry

  5. n C V (l) Worked example 1. Calculate the number of moles in 200cm3 of 0.5 mol/l sodium hydroxide solution. Step 1 :- Change the volume into litres. 0.2 litres Step 2 :- Using the triangle gives number of moles = concentration x volume (in litres). = 0.5 x 0.2 = 0.1 moles Calculations for you to try. 1. Calculate the number of moles in 50 cm3 of 0.1 mol/l zinc sulphate solution. 2. Calculate the number of moles in 0.2 litres of 2 mol/l sodium hydroxide solution Volume = 50/1000 = 0.05 litres Number of moles = 0.1 x 0.05 = 0.0005 moles Volume = 0.2 litres Number of moles = 0.2 x 2 = 0.4 moles Standard Grade Chemistry

  6. Calculations involving, concentration, moles and mass. In this type of calculation both triangles are used. n = c x V(l) = 0.1 x 0.2 = 0.02 moles n C V (l) mass n gfm mass = n x gfm = 0.02 x 40 = 0.8 g Worked example 1. Calculate the mass required to prepare 200cm3 of 0.1 mol/l sodium hydroxide, (NaOH), solution. Step 1 :- Calculate the number of moles of sodium hydroxide in 200cm3 of 0.1 mol/l of solution. Step 2 :- Calculate the mass of NaOH in 0.02 moles. gfm of NaOH = (1 x 23) + (1 x 16) + (1 x 1) = 40 Calculation for you to try. Calculate the mass of zinc sulphate in 500 cm3 of 0.2 mol/l ZnSO4(aq) Volume = 500/1000 = 0.5 litres No. of moles = 0.2 x 0.5 = 0.1 Mass = 0.1 x 161.5 = 16.15 g Standard Grade Chemistry

  7. Step 1 :- Calculate the number of moles of potassium chloride in 7.45g Number of moles = = = 0.1 mass gfm 7.45 74.5 Step 2 :- Calculate the concentration of the solution. Concentration = = = 0.4 mol/l mass n gfm n C V (l) No. of moles Volume in litres 0.1 0.25 Worked example 2. Calculate the concentration of a solution that contains 7.45 g of potassium chloride (KCl) in 250cm3 of solution. Calculation for you to try. Calculate the concentration of a solution that contains 5.85 g of sodium Chloride, NaCl, in 200cm3 of solution. Volume = 200/1000 = 0.2 litres No. of moles in 5.85 g of NaCl = 5.85/58.5 = 0.1 Concentration = 0.1 / 0.2 = 0.5 mol/l Standard Grade Chemistry

More Related