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# C2 calculations

Question | Answer |
---|---|

work out the formula mass of NaCl. Na=23; Cl=35.5 | Formula Mass of NaCl = 1xMass Na + 1xMass Cl = 23 + 35.5 =58.5 |

CHALLENGE work out the formula mass of (NH4)2SO4; N=14; H=1; S=32; O=16 | Formula mass of (NH4)2SO4= (massN x1 + mass H x4)x2 + massS x1 + massO x4= (14x1 + 1x4)x2 + 32x1 + 16x4 = 132 |

work out the formula mass of MgSO4; Mg=24; S=32; O=16 | Formula mass MgSO4 = MassMg x1 + MassS x1 + MassO x4= 24x1 + 32x1 + 16x4 = 120 |

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 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 formula mass of a compound? | multiply the mass of each element by the subscript that follows; then add the masses together; MgCl2 |

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 |

work out the formula mass of NH4NO3 | 14*1 + 1*4 + 14*1 + 16*3 = 80 |

what is the empirical formula for the compound that has 2.4g of Magnesium and 1.6g of Oxygen? Mg=24 and O=16 | Work out the moles of Mg=2.4/24=0.1 and moles of Oxygen=1.6/16=0.1; divide both numbers by the smallest: Mg=0.1/0.1=1 and O=0.1/0.1=1; the empirical formula is Mg1O1 |

what is the empirical formula for the compound that has 4.6g of Sodium and 1.6g of Oxygen? Na=23 and O=16 | Work out the moles of Na=4.6/23=0.2 and moles of Oxygen=1.6/16=0.1; divide both numbers by the smallest: Na=0.2/0.1=2 and O=0.1/0.1=1; the empirical formula is Na2O1 |

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 |

how do you work out the empirical formula, given the mass of each element? | set up one column for each element; on the first line, work out mass/RAM (ie the moles); on the second line divide the moles by the smallest number; remember to show the elements for the empirical formula (and not only the ratio) |

what is the relative atomic mass of Chlorine, knowing that it has two isotopes: Chlorine with mass 35 and an abundance of 75% and Chlorine with mass 37 and an abundance of 25% | (35*75 + 37*25)/100= 35.5 |

Boron has two isotopes of mass 10 and 11. The relative atomic mass of Boron is 10.8. What does it tell you about the abundance of each isotope? | There is more isotope of Mass 11, because the RAM is closer to 11 than 10 |

CHALLENGE 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/formula mass=80/160=0.5; ratio of BIG numbers says moles Fe= 2x molesFe2O3 =0.5x2=1; mass Fe=moles Fe x formula mass=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) | Formula mass KClO3=39+35.5+3x16=122.5; form.mass KCl=39+35.5=74.5; moles KClO3=mass/formula mass=122.5/122.5=1; ratio of BIG numbers says moles KCl= moles KClO3 =1; mass KCl=moles KCl x formula mass=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/formula mass=12.4/124=0.1; ratio of BIG numbers says moles CO2=moles CuCO3 =0.1; mass CO2=moles CO2 x formula mass=0.1x44=4.4g |

Gallium has two isotopes of mass 69 and 71. The relative atomic mass of Gallium is 69.8. What does it tell you about the abundance of each isotope? | There is more isotope of Mass 69, because the RAM is closer to 69 than 71 |

How do you work out the relative atomic mass, given the mass of each isotope and the % abundance of each isotope? | for each isotope, multiply the percentage by the mass of the isotope; add all the numbers; then divide by 100. |

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? | 3.6g (what you should collect) |

The theoretical yield of Magnesium Oxide is 4.0g. But only 3.6g is collected. What is the %yield? | 3.6/4.0*100=90% |

Created by:
UrsulineChem