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Calculations in Chemistry

- To calculate moles of a substance, use the mole triangle with the substance's mass in grams, molar mass, and moles as variables. - Examples show calculating moles of NaOH from mass and molar mass, and calculating mass of MgCl2 from moles and molar mass. - Calculating concentration involves relating moles, volume, and litres in the mole triangle. - The simplest formula for a compound can be determined by calculating moles of each element and comparing their ratios. - Neutralization calculations use moles, concentration, and volume to determine amounts of acid and base needed according reaction stoichiometry. - Calculations can also be done using a

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250 views7 pages

Calculations in Chemistry

- To calculate moles of a substance, use the mole triangle with the substance's mass in grams, molar mass, and moles as variables. - Examples show calculating moles of NaOH from mass and molar mass, and calculating mass of MgCl2 from moles and molar mass. - Calculating concentration involves relating moles, volume, and litres in the mole triangle. - The simplest formula for a compound can be determined by calculating moles of each element and comparing their ratios. - Neutralization calculations use moles, concentration, and volume to determine amounts of acid and base needed according reaction stoichiometry. - Calculations can also be done using a

Uploaded by

Khondokar Tarakky
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Calculations in Chemistry

To calculate the number of moles in a solid we use the following Mole Triangle To calculate the number of moles in a solution we use the following Mole Triangle

g n gfm c

n v

g = Mass in Grams n= Number of moles gfm=gram formula mass

n = number of moles c = concentatration (moles/litre) v= volume in litres

Examples using the Mole Triangle


Calculate the no. of moles present in 0.4g of Na OH From previous slide : If we cover up the entity we require we see that n = g/gfm
therefore

gfm of NaOH= 23+16+1= 40g

n=0.4/40 =0.01moles

Calculate the mass of 0.05 moles of Mg(Cl)2 Again from previous slide we see that if we cover up the letter we want that we get g= n gfm
therefore 0.05 95=4.75g

gfm = 24+2(35.5) =95g

Calculations contd.
Calculate the no. of moles present in 50cm3 of 0.05 molar HCl . From previous triangle we that if we cover the letter we want that n =cv/1000 therefore n= 0.0550/1000= 0.00005moles

Calculate the concentration if we have 0.1 moles dissolved in 100cm3 of water From previous triangle we see that c= n/v in litres
therefore

C = 0.1/100/1000= 0.01 moles/litre

Empirical or Simplest formula


Example: A sample of a substance was found to contain 0.12g of Magnesium and 0.19g of Fluorine. Find the simplest Formula. Rules 1. Write down all the symbols present. 2. Calculate the no of moles of each element present. 3. Compare ratios(get the smallest number of moles and divide it into all the others. 4.Write down the formula. Mg and F n =g/gfm
0.12/24 = 0.005moles

It doesnt matter if the original sample is in grams or percentages

0.19/19 = 0.01moles

0.005 : 0.01 1: 2 Mg(F)2

Neutralisation Calculations
One way of Neutralising an Acid is to add an Alkali(for other methods see reactions of acids
section).ie. Acid + Alkali Salt + water To do neutralisation calculations we use the following formula. H+ CA VA = OH- CB VB acid alkali

OH- = no. of OHH+ = no. of H+ ions CA =Concentration of acid ions in alkali in acid HCl = 1 NaOH = 1 CB= Concentration of alkali H2SO4 = 2 Ba(OH)2 = 2 VA = volume of acid H3PO4 = 3 Al(OH)3 = 3 VB =Volume of alkali

Neutralisation Calculations contd.


Example: What volume of 0.1M HCl is required to neutralise 100cm3 of 0.5M NaOH. H+ CA VA = OH- CB VB
We require to find

1 0.1 VA = 1 0.5 100 VA = 50/0.1 = 500cm3

Example:What concentration of 50cm3 KOH is used to neutralise 100cm3 of 0.05M H2SO4 H+ CA VA = OH- CB VB
We require to find

2 0.05 100 = 1 CB 50 CB = 10/50 = 0.2M

Calculations from Equations


Example: What mass of Hydrogen gas is produced when 0.12g of Magnesium is added to excess Hydrochloric Acid.
We first require to write down the balanced equation

Mg(s) + 2 HCl(aq)
From This we can see that:

Mg(Cl)2(aq) + H2(g)

To balance the equation we add a 2 in front of the HCl 1M 24g 1g


therefore

1M

1M

2g 2/24 =0.08 0.080.12 = 0.0096g of H2

0.12g

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