anonymous
  • anonymous
(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
katieb
  • katieb
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anonymous
  • anonymous
@gerryliyana
anonymous
  • anonymous
(Ar Nitrogen/Mr NaNo3) x mass of NaNO3
anonymous
  • anonymous
Ar stands for what?

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anonymous
  • anonymous
mass of relative atom
anonymous
  • anonymous
Mr is "mass of relative molecule"
anonymous
  • anonymous
ok
anonymous
  • anonymous
no need to times 100g?
anonymous
  • anonymous
do you have a periodic table?
anonymous
  • anonymous
yes
anonymous
  • anonymous
wait a moment
sirm3d
  • sirm3d
\[\large m \text { N}=100g \text { Na} \times \frac{ 14g \text{ N} }{ 23g \text { Na} } \approx 61g \text { N}\]
anonymous
  • anonymous
Na=23.0g N=14.0g O=16.0g
anonymous
  • anonymous
why 23g Na?? not NaNO3?
anonymous
  • anonymous
but my idea is this... |dw:1354705910345:dw|
sirm3d
  • sirm3d
our interest is in the element Na and N in the compound. the ratio of these elements is 1:1
anonymous
  • anonymous
uhhh...what do you mean?
sirm3d
  • sirm3d
the compound contains 100g of Na. it did not say the compound NaNO3 weighs 100g.
anonymous
  • anonymous
yes sodium is Na.., so 100 is mass of Na
anonymous
  • anonymous
@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?
sirm3d
  • sirm3d
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.
anonymous
  • anonymous
represent mass of Na
anonymous
  • anonymous
@sirm3d okay, how about part b? @gerryliyana thanks, I got it :) part b please
anonymous
  • anonymous
4.6*(10H2O/Na2) ?
sirm3d
  • sirm3d
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
sirm3d
  • sirm3d
mass of 10 moles of water is 10 x 18g = 180 g
sirm3d
  • sirm3d
we convert 4.6 g Na to g H2O by the mass ratio in the compound (180 g H2O / 23 g Na)
anonymous
  • anonymous
but there are two Na atom.... Na2Co3
anonymous
  • anonymous
sorry for typing mistakes, it should be Na2CO3 10H2O
sirm3d
  • sirm3d
sorry, missed that two Na. Mass of Na in compound is 46 g. use the mass ratio 180g H2O / 46g Na
anonymous
  • anonymous
|dw:1354706692310:dw|
sirm3d
  • sirm3d
right.
anonymous
  • anonymous
thanks. How about part c?
sirm3d
  • sirm3d
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.
anonymous
  • anonymous
I got this one
sirm3d
  • sirm3d
i got 63.5g M
anonymous
  • anonymous
so what is the formula about finding relative atomic mass of M?
anonymous
  • anonymous
there is no answer on my textbook. I cannot ask my teacher since I will have Chemistry examination tonight.
sirm3d
  • sirm3d
it is the mass of M relative to the mass of O
anonymous
  • anonymous
there are 1 atom of M, right?
sirm3d
  • sirm3d
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
anonymous
  • anonymous
so M is copper....
anonymous
  • anonymous
mass of M=79.87g and the relative atomic mass is ...?
sirm3d
  • sirm3d
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.
anonymous
  • anonymous
how can you get the answer?
sirm3d
  • sirm3d
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.
sirm3d
  • sirm3d
you cannot give the actual mass of M in the compound MO if you do not have the actual mass of MO.
sirm3d
  • sirm3d
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?
anonymous
  • anonymous
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...
sirm3d
  • sirm3d
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.
sirm3d
  • sirm3d
16.0g O x 79.87g M / 20.13g O = 63.5g M
anonymous
  • anonymous
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?
sirm3d
  • sirm3d
yep. that's how you convert the given.
anonymous
  • anonymous
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
sirm3d
  • sirm3d
|dw:1354708533879:dw| observe how i got rid of g O and change it to g M
anonymous
  • anonymous
I know it is the answer of part d 168 g M?
anonymous
  • anonymous
are you here?
sirm3d
  • sirm3d
1 got 132.32 g M
anonymous
  • anonymous
ummm......what are your steps?
sirm3d
  • sirm3d
g M in MCl = 268.88g-5.68g=21.2g M
sirm3d
  • sirm3d
mass ratio is 21.2g M / 5.68g Cl
anonymous
  • anonymous
ummmm...you get something wrong.... is 26.88g instead of 268.88g
sirm3d
  • sirm3d
sorry. typo error.
sirm3d
  • sirm3d
relative mass of Cl = 35.453g
anonymous
  • anonymous
just take the relative mass of Cl =35.5g
sirm3d
  • sirm3d
35.5g Cl x 21.2g M / 5.68g Cl = 132.5g M
sirm3d
  • sirm3d
M = Cs
anonymous
  • anonymous
why 21.2 g?
anonymous
  • anonymous
26.88g-5.68g?
anonymous
  • anonymous
okay I got it. Thank you very much. :)
sirm3d
  • sirm3d
yep. 26.88g - 5.68g = 21.2g
anonymous
  • anonymous
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.
sirm3d
  • sirm3d
hmm. 0.213 g O2 used in the process.

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