Chemistry trouble (1 Viewer)

[st1m]

Hey I'm having a bit of trouble understanding Mass-mass calculations.

Anybody help? Notes, texts? conquering chemistry confuses me because my teacher explained the concept in a different way.
bye

 

[Aerath]

What about mass-mass calculations? There are a lot of good quality notes in the resource section of BoS.

 

[bored of sc]

The basic equations are...

Moles of a monatomic (not diatomic or polyatomic e.g. can't be H2, S8 etc) element = [mass of substance (in grams)] / [Ar (the relative atomic mass, see periodic table for atomic mass number)]

Moles of a diatomic/polyatomic element/compound (e.g. H2O, P9 etc) = [mass of substances (in grams)] / [Mr (the relative molecular mass, see periodic table atomic mass number of atoms and add together)]

Moles of gases = [volume of gas (dm^3)] / [24.79 (dm^3)]

Example: What mass of magnesium oxide will be produced if 1.5cm (0.12g) of Mg ribbon is burned?

Equation: 2Mg (s) + O2 (g) --> 2MgO (s), therefore the ratio of moles of Mg to O2 to MgO is 2:1:2 (number in front of atomic symbol, note: if no symbol is written there the number is 1)

Relative atomic mass (Ar) of Mg = 24, relative atomic mass of oxygen = 16 (see periodic table mass numbers)

Number of moles of Magnesium = mass Mg (grams) / Ar Mg = 0.12/24 = 0.005 moles

Therefore the number of moles of MgO = 0.005 (since 0.005 is number of moles of Mg and the ratio of moles of Mg:O:MgO = 2:1:2 the number of moles of MgO in this case is equal to the number of moles of Mg (which is 2, the numbers in front of the atomic symbols)

Number of moles of MgO = mass MgO / Mr (relative molecular mass) MgO

Substitute in values of equation you know/can find out via periodic table therefore...

0.005 (number of moles of MgO, as established above) = mass MgO / [(24 + 16) relative molecular masses (periodic table values) of Mg + (added with) O]

0.005 = mass MgO (grams) / 40

Solve like simple linear equation (in above case multiply both sides by 40)

Thus the mass MgO = 0.005 * 40 = 0.2 grams

Ratios: the ratio of moles changes (but is equivalent to) the simple ratios established through examining the equation (i.e. the numbers preceding the atomic symbols) e.g. as above, the number of moles of Mg was established to be 0.005. The ratio of moles of Mg to O to MgO was 2:1:2, therefore change Mg's ratio number to 0.005 > therefore you must change the other numbers in the ratio too (to ensure it is equivalent), since 2 has changed to 0.005 for Mg (first number of ratio), the 1 for oxygen changes of 0.0025 (half of 0.005 since 1 is half of 2) and the 2 also changes to 0.005 since the ratios must stay equivalent. Note: If you were to simplify 0.005:0.0025:0.005 to the simpliest whole number ratio (i.e. empirical formula you would get 2:1:2).

Further hints - work with what you HAVE been given, don't make it harder for yourself. In the above case you are given the mass of Mg (as 0.12g) so work with the mass of Mg to find the number of moles of Mg - then your answer should seem fairly straight forward.
- show ALL working and explanations of working, if you are in doubt as to detail, write down basically EVERYTHING you used in the development of your answer in a clear, logical way
- don't forget to multiply the relative molecular masses of compounds like H2O (i.e. Hydrogen would be 2 times the relative molecular mass (as in periodic table) since there are 2 hydrogen atoms in the simpliest molecule
- never lack confidence (don't go with the wrong attitude of 'I can't do this') because if you have that attitude the answer seems harder than it is to find)

For possibly a better explanation send me a PM, or nag anyone smart (i.e. Lynonamu, Aerath, Razizi, Kaz, Jelly Belly, the chick from the Gold Coast with the colourful picture (lol) ... anyone really who does Chemistry)

 

[bored of sc]

If someone has a special aptitude/ability in Chemistry and has good, clear explanation skills, your help would be much appreciated.

 

[st1m]

thanks bored of sc

i will have a look at the stuff you wrote later...

thnx again

 

[bored of sc]

thanks bored of sc

i will have a look at the stuff you wrote later...

thnx again

no worries

 

[123huss123]

i wanna murder
chemistry
**random**

 

[Continuum]

You don't really need to remember all the formulae for moles... as long as you know the concept, most of the stuff is just logic.

 

[selablad]

You don't really need to remember all the formulae for moles... as long as you know the concept, most of the stuff is just logic.

:uhhuh: I find relying on logic is easier than formulae anyway