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AQA A1 Calculations
AQA calculations
| Question | Answer |
|---|---|
| What are isotopes | Atoms of the same element that have the same number of PROTONS (atomic) but different number of NEUTRONS (mass) |
| What is the standard for Relative Mass, to which all other masses of atoms are compared to? | Carbon-12 |
| What is the relative Isotopic Mass? | the mass of an atom of an isotope, compared with 1/12th of the mass of an atom of Carbon-12 |
| What is the Relative Atomic Mass (Ar)? | the mean mass of an atom weighed according to the abundance of the isotopes (and compared with 1/12th of the mass of an atom of Carbon-12) |
| What is the Relative Molecular Mass? | the mean mass of a molecule weighed according to the abundance of the isotopes of each element(and compared with 1/12th of the mass of an atom of Carbon-12) |
| What is the Relative Formula Mass? | The same as the RMM, but applied to giant structure (that do not exist as molecules!) |
| What is the RAM of Boron, when there is 19.77% of Boron-10, 80.23% of Boron-11? (give to 4sf) | 10.80 |
| Work out the RAM of Chlorine knowing that there are only two isotopes Cl-35 and Cl-37 and that there is 3 times more Cl-35 than Cl-37 | Set up the equation: RAM= (sum of abundancexmass) / total abundance; here there are 3 parts Cl-35 to 1 part Cl-37 so total abundance is 4; RAM=(35*3+37*1)/4=35.5 |
| The RAM of Silicon is 28.11 a.u. . what does it mean in terms of abundance of isotopes? | There is more atoms of isotope-28 than there is of any other isotopes. |
| The RAM of Rubidium is 85.47; there are only two isotopes: Rb-85 and Rb-87 ;work out the abundance of Rb-85 | Set up the equation: RAM= (sum of abundancexmass) / total abundance =85.47=[85*x+87*(100-x)]/100; then solve the equation; x=76.5 |
| What does Avogadro's constant represent? | The number of atoms in a mole of Carbon-12 |
| What is the value of Avogadro's number? | 6.02x10^23 atoms per mole |
| What is the Molar Mass? | The mass of a mole of a substance. |
| What is the unit of Molar Mass? | g.mol-1 |
| How do you work out the amount of substance in a given mass? | moles=mass/Molar Mass |
| How do you work out the mass of a substance, given the amount? | mass=moles x Molar Mass |
| What is the Empirical Formula? | The simplest whole number ratio of atoms of each element present in a compound |
| What is the Molecular Formula? | The number of atoms of each element that make up a molecule |
| How do you work out the Empirical formula from masses of each element? | 1) Divide the mass of each element by the RAM; 2)divide by the smallest number 3) Check for SPECIAL ratio is whole numbers (so check for .25, .5, .33 or .66) |
| What is the empirical formula of a compound that has H 3.66%; P 37.8%; O 58.5%? | 1)moles H=3.66/1=3.66; P=37.8/31.0=1.219 O=58.5/16=3.656; 2)divide by smallest H=3.66/1.219=3.00 P=1 O=3.656/1.219=3; 3)Special ratio? Not here so empirical is H3PO3 |
| What is the empirical formula of a compound that has Pb 90.66%; O 9.34%? | 1)moles Pb=90.66/207.2=0.437; O 9.34/16=0.584; 2)divide by smallest Pb=1; O=0.584/0.437=1.336; 3)Special ratio? Yes (.33 so multiply by 3) Empirical formula is Pb3O4 |
| How do you work out the Molecular Formula from the masses of each element and the RMM? | 1)Work out the empirical formula; 2)Work out the mass of the empirical formula; 3)work out the number of units in the molecular formula by dividing RMM by mass-of-empirical-formula; 4) Molecular formula is the number of units x empirical formula |
| What is Avogadro's Law? | equal volumes of gases at the same temperature and pressure contain the same number of moles |
| How do you work out the amount of gas in a given volume? | Number of moles=(PxV)/(RxT). Have you remembered to convert cm3 into m3? Celsius into Kelvins? Atm into Pa? |
| How do you work out the volume of gas, given the number of moles? | Volume (in m3) = (nxRxT)/P. Have you remembered to convert cm3 into m3? Celsius into Kelvins? Atm into Pa? |
| How do you convert cm3 into dm3? | Divide by 1000 |
| How do you convert cm3 into m3? | Divide by 1,000,000 |
| How do you convert dm3 into cm3? | Multiply by 1000 |
| How do you convert m3 into cm3? | Multiply by 1,000,000 |
| What is the number of moles in 1080dm3 at 25C and 100kPa? | Moles = (1.080m3x100,000Pa)/(8.31*298K) = 43.61 (did you remember to convert? Have you checked the significant figures?) |
| What is the volume of 0.25moles of Oxygen gas? | Volume = 0.25 *24 = 6dm3; Have you remembered the units? |
| What is the mass in mg of 0.60dm3 of Nitrogen gas (N2) at room temp and 100kPa? | Moles = (0.00060*8.31*298) / 100,000 = 1.48x10^-5; mass=molesxRMM=1.48x10^-5*28=0.41mg; Have you remembered to convert g into mg? |
| What is the volume of 1.282g of SO2 gas? | Moles = mass/RMM=1.282 / 64.1= 0.02; volume=(nxRxT)/P = (0.02 x8.31x298)/100,000 = 0.495dm3; Have you remembered the units? |
| What is the mass of 0.500 moles of Fe2O3? | RMM of Fe2O3 is 2x55.8+3x16.0=159.6 g.mol-1; Mass = 0.500 x 159.6 = 79.8g; Have you remembered the units? And significant figures? |
| What is the amount of Al2(SO4)3 in 10.269g? | RMM of Al2(SO4)3=27.0x2+(32.1+4x16)x3=342.3 g.mol-1; Moles = 10.269 / 342.3 = 0.0300; Have you remembered the units? And significant figures? |
| What is the concentration of a solution? | The amount of solute dissolved in 1dm3 of solvent |
| What is a standard solution? | A solution of known concentration |
| How do you make up a standard solution? | 1)weigh out the solute by difference;2)dissolve ina beaker and transfer into flask;3)rinse beaker, funnel and rod into flask;4)top up so that BOTTOM of meniscus is on graduation line |
| What is a molar solution? | A solution with a concentration of 1 mole.dm-3 |
| What is the unit of concentration? | mol.dm-3 |
| What is the amount of solution in 24.35cm3 of a 0.125M solution of Potassium Nitrate KNO3? | moles = 24.35/1000 x 0.125 = 0.00304 or 3.04x10^-3 |
| What is the concentration of 8.75x10^-3 moles in 50cm3 of solution? | Concentration = 8.75x10^-3 / (50/1000)= 0.175 mole.dm-3; Have you remembered to convert cm3 into dm3?; Have you remembered the units? |
| What is the mass concentration of 3.56x10^-3 moles of H2SO4 dissolved in 25cm3 of solution? | RMM H2SO4=2x1.0+32.1+4x16.0=98.1; Mass= 3.56x10^-3 x 98.1 = 0.349g; Concentration= 0.349 / (25/1000) = 13.96 g.dm-3; Have you remembered the units at each stage? |
| Al=27; O=16 | nb of atoms of Aluminium x mass of Aluminium x 100 / Formula mass of Al2O3 = 2 x 27 x 100 /102 = 57% |
| work out the % by mass of Oxygen in Magnesium Sulfate, MgSO4; Mg=24; S=32; O=16 | nb of atoms of Oxygen x mass of Oxygen x 100 / Formula mass of MgSO4 = 53.3% |
| work out the % by mass of Aluminium in Aluminium Oxide, Al2O3; Al=27; O=16 | nb of atoms of Aluminium x mass of Aluminium x 100 / Formula mass of Al2O3 = 2 x 27 x100 / 102 = 57% |
| how do you work out the % by mass of an Element in a compound | Work out the formula mass; then check the number of atom of your element; %by mass= number of atoms x RAM of element x 100 / formula mass |
| CHALLENGE: 239g of a Lead Oxide contains 32g of Oxygen. what is the empirical formula? Pb=207; O=16 | You must have the mass of each ELEMENT: here mass of Lead is 239-32=207; Then as usual, Work out the moles of Pb=207/207=1 and moles of Oxygen=32/16=2; divide both numbers by the smallest: Pb=1/1=1 and O=2/1=2; the empirical formula is Pb1O2 |
| Fe2O3 + 3 CO -> 2 Fe + 3 CO2; what mass of iron is produced when 80g of Iron(III)oxide is reacted? (Fe=56; O=16) | Formula mass Fe2O3=56x2+3x16=160; moles Fe2O3=mass/RMM=80/160=0.5; ratio of BIG numbers says moles Fe= 2x molesFe2O3 =0.5x2=1; mass Fe=moles Fe x RMM=1x56=56g; remember to use BIG numbers in ratio ONLY |
| 2 KClO3 -> 2 KCl + 3 O2; What mass of KCl is produced when 122.5g of KClO3 reacts completely? (K=39; Cl=35.5; O=16) | RMM KClO3=39+35.5+3x16=122.5; form.mass KCl=39+35.5=74.5; moles KClO3=mass/RMM=122.5/122.5=1; ratio of BIG numbers says moles KCl= moles KClO3 =1; mass KCl=moles KCl x RMM=1x74.5=74.5g |
| CuCO3 -> CuO + CO2; What mass of Carbon Dioxide is produced when 12.4g of CuCO3 reacts completely? (Cu=64; C=12; O=16) | Formula mass CuCO3=64+12+3x16=124; form mass CO2=12+16x2=44; moles CuCO3=mass/RMM=12.4/124=0.1; ratio of BIG numbers says moles CO2=moles CuCO3 =0.1; mass CO2=moles CO2 x RMM=0.1x44=4.4g |
| how do you work the mass of a compound, given the equation for the reaction and the mass of the reactant? | set up the calculations under the equation: first line is mass; second line is formula mass; bottom line is moles; then go down (divide mass by formula mass); across (look at the BIG numbers); and up (multiply moles by formula mass) |
| In an experiment 2.4g of Magnesium is burned; and 3.6g of Magnesium Oxide is collected. What is the %yield of Magnesium Oxide? | molesMg=2.4/24=0.1; ratio: molesMgO=molesMg; THEORETICAL mass of MgO=moles MgOxRMM=0.1*(24+16)=4.0. But only 3.6g is collected. What is the %yield? |
| A known mass of the hydrated salt, MCl2.6H2O was heated to constant mass. Mass of hydrated salt=5.51 g; Mass of anhydrous salt =3.01 g. Use the results to calculate the Mr of the anhydrous salt. Calculate the RAM and deduce the identity of the metal | Mass H2O=2.5g; moles H2O=0.139; Moles MCl2=molesH2O/6=0.0231; RMM=5.51/0.0231=130;130=RAM+71; RAM=59; Metal=cobalt |
| A known mass of a hydrated carbonate X2CO3.10H2O was heated in a crucible to constant mass, Mass of hydrated carbonate =2.01 g; Mass anhydrous carbonate =0.74 g Identify the metal X. | MassH2O=1.27g;MolesH2O=0.0705;moles anhydrous=0.0705/10=0.00705; RMM=mass/moles=0.74/0.00705=105.96=2*RAM+60;RAM=23;Metal=Na |
| 1.6 g of methane was reacted with chlorine to make chloromethane. Cl2 + CH4 --> CH3Cl + HCl. The yield is 59%. What mass of product was collected? | 1) Work out theoretical mass of CH3Cl=1.6/16*50.5=5.05g;2) mass collected = %yield * theoretical = .59*5.05=2.97g |
| the % yield in the following reaction is 80.0%. Mg + 0.5 O2 --> MgO. How much magnesium needs to be burned to make 30.0g of magnesium oxide? | 1) Work out theoretical mass from %yield: theoretical=collected/yield=30/.8=37.5; 2)Down-across-Up to work out Mass reactant: Mass=37.5/40*24=22.5g |
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