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(a) What is the mass of nitrogen present in the sample of sodium nitrate (NaNO3) which contains 100g of sodium? (b) What is the mass of water of crystallization present in the sample of sodium carbonate-10-water (Ma2CO3·10H2O) which contains 4.6g of sodium? (c) A metal oxide MO contains 79.87% by mass of the metal M. Find the relative atomic mass of M. (d) 26.88g of a metal chloride MCl contains 5.68 g of chlorine. Find the relative atomic mass of the metal M.

Chemistry
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(Ar Nitrogen/Mr NaNo3) x mass of NaNO3
Ar stands for what?

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Other answers:

mass of relative atom
Mr is "mass of relative molecule"
ok
no need to times 100g?
do you have a periodic table?
yes
wait a moment
\[\large m \text { N}=100g \text { Na} \times \frac{ 14g \text{ N} }{ 23g \text { Na} } \approx 61g \text { N}\]
Na=23.0g N=14.0g O=16.0g
why 23g Na?? not NaNO3?
but my idea is this... |dw:1354705910345:dw|
our interest is in the element Na and N in the compound. the ratio of these elements is 1:1
uhhh...what do you mean?
the compound contains 100g of Na. it did not say the compound NaNO3 weighs 100g.
yes sodium is Na.., so 100 is mass of Na
@sirm3d so it depends on the question asking about the mass of something which contains how many grams of that thing, right? @gerryliyana that hundred of sodium represents what? the mass or the weight?
yes. be careful of the question. if the 100g is the mass of the compound NaNO3, then we need to compute the mass on Na in the compound.
represent mass of Na
@sirm3d okay, how about part b? @gerryliyana thanks, I got it :) part b please
4.6*(10H2O/Na2) ?
same thing. The mass of Na in the compound, sodium carbonate decahydrate, is 23g. we compute the total mass of water molecules in the compound
mass of 10 moles of water is 10 x 18g = 180 g
we convert 4.6 g Na to g H2O by the mass ratio in the compound (180 g H2O / 23 g Na)
but there are two Na atom.... Na2Co3
sorry for typing mistakes, it should be Na2CO3 10H2O
sorry, missed that two Na. Mass of Na in compound is 46 g. use the mass ratio 180g H2O / 46g Na
|dw:1354706692310:dw|
right.
thanks. How about part c?
assume 100g mass of compound MO. Mass of M in 100g compound is 79.87 g, mass of O in 100g compound is 20.13 g.
I got this one
i got 63.5g M
so what is the formula about finding relative atomic mass of M?
there is no answer on my textbook. I cannot ask my teacher since I will have Chemistry examination tonight.
it is the mass of M relative to the mass of O
there are 1 atom of M, right?
yes, 1:1 ratio. so the mass of the oxygen in MO is 16g. the mass of M relative to 16g O present in the compound is the relative mass of M
so M is copper....
mass of M=79.87g and the relative atomic mass is ...?
if there is 1 mol of MO, the mass of M is 63.5 g. if there are 2 mol of MO, the mass of M is 127g. relative mass means mass in ONE mol of the compound.
how can you get the answer?
we assumed that IF there are 100g of MO, clearly 79.77% of that mass belongs to M. you would arrive at the same result if you assume that the starting mass of MO is 20g. In this 20g MO, the mass of M is 79.87% of 20g while 20.13% of 20g is the mass of O.
you cannot give the actual mass of M in the compound MO if you do not have the actual mass of MO.
i'll give an analogy just to be clear with the "relative" mass. Suppose you got 90% of the questions in your exam, can you tell me how many correct answers you got?
if there were 100 questions in my exam, I got 90 questions correct. I know why you assume 100 g be the mass of the compound MO and the ratio is 1:1 total mass =100 g mass of M=79.87 g mass of O=20.13 g relative atomic mass of O=16.0 g But I am confusing how to get the relative atomic mass of M...... I still cannot get you idea about M...
we convert the relative atomic mass of O = 16.0g to relative atomic mass of M using the available masses of 79.87g M and 20.13g O.
16.0g O x 79.87g M / 20.13g O = 63.5g M
the relative atomic mass of O is known and ask for the relative atomic mass of M, we need to make the mass of O be the denominator and mass of M be the numerator. If the relative atomic mass of M is known and ask for the relative atomic mass of O, we need to make the mass of M be the denominator and mass of O be the numerator. is it?
yep. that's how you convert the given.
thank you for correcting my concepts and I am working with part d, I will give you response when I get my answer and you check is my answer correct or not
|dw:1354708533879:dw| observe how i got rid of g O and change it to g M
I know it is the answer of part d 168 g M?
are you here?
1 got 132.32 g M
ummm......what are your steps?
g M in MCl = 268.88g-5.68g=21.2g M
mass ratio is 21.2g M / 5.68g Cl
ummmm...you get something wrong.... is 26.88g instead of 268.88g
sorry. typo error.
relative mass of Cl = 35.453g
just take the relative mass of Cl =35.5g
35.5g Cl x 21.2g M / 5.68g Cl = 132.5g M
M = Cs
why 21.2 g?
26.88g-5.68g?
okay I got it. Thank you very much. :)
yep. 26.88g - 5.68g = 21.2g
1.200 g of a compound containing only C, H and O gave 1.173 g of CO2 and 0.240 g of H2O on complete combustion. Find the empirical formula of the compound.
hmm. 0.213 g O2 used in the process.

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