1.34C: Understand How to Carry Out Calculations involving Amount of Substance, Volume and Concentration (in Mol / dm3 ) of Solution

CALCULATIONS INVOLVING SUBSTANCES

GENERAL EQUATION:

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1. CALCULATING MOLES

EQUATION:

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Amount of Substance (Mol)    =   Concentration     x     Volume of Solution (dm3)

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Example:

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Calculate the Moles of Solute dissolved in 2 dm3 of a 0.1 mol / dm3 Solution

Concentration of Solution : 0.1 mol / dm3 Volume of Solution : 2 dm Moles of Solute   =   0.1   x   2   =   0.1                                                                                          Amount of Solute = 0.2 Mol

    2.  CALCULATING CONCENTRATION

EQUATION:

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Example:

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25.0 cm3 of 0.050 mol / dm3 Sodium carbonate were completely Neutralised by 20.00 cm3 of Dilute Hydrochloric Acid. Calculate the Concentration, in Mol / dm3 of the Hydrochloric Acid

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STEP 1 - Calculate the Amount, in Moles, of Sodium Carbonate Reacted                1 dm3 = 1000 cm3                Amount of Na2CO3  =  ( 25.0 ÷ 1000) x 0.050  =  0.00125 Mol STEP 2 - Calculate the Amount, in Moles, of Hydrochloric Acid Reacted                Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2                1 Mol of Na2CO3 Reacts with 2 Mol of HCl                0.00125 Mol of  Na2CO3 Reacts with 0.00250 Mol of HCl STEP 3 - Calculate the Concentration, in Mol / dm3, of the Hydrochloric Acid                1 dm3 = 1000 cm3                Concentration ( Mol / dm3 ) =  0.00250 ÷ ( 20 ÷ 1000 )  =   0.125                                            Concentration of Hydrochloric Acid = 0.125 Mol / dm3

    3.  CALCULATING VOLUME

EQUATION:

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Example:

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Calculate the Volume of Hydrochloric Acid of Concentration 1.0 mol / dm3 that is required to react completely with 2.5g of Calcium Carbonate

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STEP 1 - Calculate the Amount, in Moles, of Calcium Carbonate that Reacts                Mr of CaCO3 is 100                Amount of CaCO3  =  ( 2.5 ÷ 100 )  =  0.025 Mol STEP 2 - Calculate the Moles of Hydrochloric Acid Required                CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2                1 Mol of CaCO3 Requires with 2 Mol of HCl                0.025 Mol of  CaCO3 Requires 0.05 Mol of HCL STEP 3 - Calculate the Volume of HCl Required                Volume  =  ( Mol of Substance  ÷  Concentration )                              =  0.05  ÷  1.0                              =  0.05 Mol                                                                   Volume of Hydrochloric Acid = 0.05 Mol