The Mole Concept Flashcards
Define the term relative atomic mass (Ar)
the average mass of one atom of an element compared with 1/12 the mass of a carbon-12 atom
The mass of carbon-12 is exactly ____ units
12.00
State the Ar (relative atomic mass) of the following:
- Cl
- Na
- H
- Ca
- S
- C
1) 35.5
2) 23
3) 1
4) 40
5) 32
6) 12
The Ar/RAM is located at the ___ of the element
top
Define relative molecular mass (Mr)
the average mass of one molecule of a substance compared to 1/12 the mass of a carbon-12 atom
Mr (H2O)
= (2*1) +16
= 2 + 16
= 18
Mr (C6H1206)
= (612) + (121) + (6*16)
= 72 + 12+ 96
= 180
The relative formula is the same as relative molecular mass but applies to ____ only
ions
Define relative formula mass
the total of Ar of all atoms in the formula of an ionic compound
What is the relative formula mass of magnesium sulphate (MgSO4)?
Mr(MgSO4) = 24 + 32 + (4*16)
= 24 + 32 + 64
= 120
Define mole
the amount of a substance that contains 6.02 *10 ^23 particles of the substance
12.00 grams of carbon-12 contains _______ carbon atoms
6.023*10^23
“Avogadro number” or “Avogadro constant” is _______
6.023 * 10 ^23
When counting ‘particles’ the unit used is the _____
mole
Calculate the number of moles in 80g of Magnesium (Mg)
24 grams of Mg contains one mole
24= Mr
; 80 grams of Mg contains 80*1/24 = 3.33 moles
OR
n=mass/molar mass
= 80/24
= 3.33 moles
Calculate the no. of moles in 24g of Magnesium sulfate (MgSO4)
Mr(MgSO4)= 24 + 32 + (4*16)
= 120
;120 g of MgSO4 contains 1 mole
24g = 24 *1/120
= 0.2 moles
OR
n= mass/ Mr
= 24/120
= 0.2 moles
Calculate no. moles in 18.06 *10^23 molecules of chlorine gas (Cl2)
6.02*10^23 molecules of gas are present in 1 mole of Cl
;18.06 * 10^23 molecules of gas are present in
18.06 *10 ^23 *1 / 6.02 *10^23
= 3 moles of Cl
Calculate the mass of 2.6 moles of Na2CO3
Mr (Na2CO3) = (223) + 12 + (163)
= 106g
1 mol( Na2CO3)= 106g
;2.6 moles of Na2CO3 =
2.6*106/1= 275.6g
Or
mass= g*no. mol
= 106 *2.6
= 275.6 g
Calculate the no of molecules in 2.1 moles of H2 gas
1 mole of H2 gas contains 6.0210^23 molecules
;2.1 moles of H2 gas contains 2.16.02*10^23/ 1
= 1.264 *10^24 molecules
State Avogadro’s law states that
equal volumes of all gases at the same temperature and pressure contain the same number of molecules
Define molar volume
the volume occupied by one mole of gas
State the standard temperature and pressure (stp) in the volume of 1 mole of gas
22.4dm^3
State the room temperature and pressure (rtp) in the volume of 1 mole of gas
24dm^3
Calculate the volume occupied by 0.2 moles of CO2 at rtp
1mole of CO2 at rtp occupies 24dm^3
; 0.2 moles of CO2 at rtp occupies 24*0.2/1 = 4.8dm^3
Calculate the number of moles in 1.3 dm^3 of NH3 at stp
22.4 dm3 of NH3 contains 1 mole
; 1.2 dm3 of NH3 contains 1.2*1/22.4=0.054 moles
Calculate the volume occupied by 34g of NH3 at rtp
Mr NH3 = 14 + (31) = 17
; the mass of 1 mole of NH3 is 17g
17g of NH3 occupies 24 dm^3 at rtp
;34g of NH3 occupies 3424/17 =48 dm3 at rtp
The concentration of solution can be expressed in
mol per dm^3 (mol dm^-3)
gram per dm^3 (g dm^-3)
1dm^3 = _____ cm^3
1000
Calculate the concentration in mol dm^-3 in 60g of NaOH in 750cm^3 of the soulution
Mr(NaOH) = 23 + 16 + 1 = 40
; the mass of 1 mole of NaOH is 40 g
40g of NaOH contains 1 mole
; 60g of NaOH contains 60*1/40 = 1.5 moles
750cm^3 contains 1.5 moles
; 1000cm^3 contains 1.5*1000/750 = 2.0 moles of NaOH
The concentration of the solution is therefore 2.0 mol dm^-3
Calculate the concentration in mol dm^-3 in 80g of Na2CO3 in 800cm^3 of the soulution
Mr(Na2CO3) = (23 * 2) + 12 + (16*3) = 106
106g contains 1 mole of Na2CO3
; 80 g of Na2CO3 = 80*1/106 =0.755
800cm^3 contains 0.755 mols
1000cm^3 contains 1000*0.755/800= 0.943mol/dm3
The concentration of the solution therefore is 0.943mol dm^-3
Calculate the percentage of Nitrogen present in the fertilizer Ammonium Nitrate (NH4NO3)
mass of 1 mole NH4NO3= 14 + (41) + 14 + (163)= 80g
Mass of Nitrogen present in 1 mole of NH4NO3= 14+14= 28g
% of Nitrogen present in NH4NO3 fertilizer
= 28/80 *100
= 35%
