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## anonymous 3 years ago (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.

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1. anonymous

@gerryliyana

2. anonymous

(Ar Nitrogen/Mr NaNo3) x mass of NaNO3

3. anonymous

Ar stands for what?

4. anonymous

mass of relative atom

5. anonymous

Mr is "mass of relative molecule"

6. anonymous

ok

7. anonymous

no need to times 100g?

8. anonymous

do you have a periodic table?

9. anonymous

yes

10. anonymous

wait a moment

11. sirm3d

$\large m \text { N}=100g \text { Na} \times \frac{ 14g \text{ N} }{ 23g \text { Na} } \approx 61g \text { N}$

12. anonymous

Na=23.0g N=14.0g O=16.0g

13. anonymous

why 23g Na?? not NaNO3?

14. anonymous

but my idea is this... |dw:1354705910345:dw|

15. sirm3d

our interest is in the element Na and N in the compound. the ratio of these elements is 1:1

16. anonymous

uhhh...what do you mean?

17. sirm3d

the compound contains 100g of Na. it did not say the compound NaNO3 weighs 100g.

18. anonymous

yes sodium is Na.., so 100 is mass of Na

19. 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?

20. 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.

21. anonymous

represent mass of Na

22. anonymous

@sirm3d okay, how about part b? @gerryliyana thanks, I got it :) part b please

23. anonymous

4.6*(10H2O/Na2) ?

24. 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

25. sirm3d

mass of 10 moles of water is 10 x 18g = 180 g

26. sirm3d

we convert 4.6 g Na to g H2O by the mass ratio in the compound (180 g H2O / 23 g Na)

27. anonymous

but there are two Na atom.... Na2Co3

28. anonymous

sorry for typing mistakes, it should be Na2CO3 10H2O

29. sirm3d

sorry, missed that two Na. Mass of Na in compound is 46 g. use the mass ratio 180g H2O / 46g Na

30. anonymous

|dw:1354706692310:dw|

31. sirm3d

right.

32. anonymous

thanks. How about part c?

33. 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.

34. anonymous

I got this one

35. sirm3d

i got 63.5g M

36. anonymous

so what is the formula about finding relative atomic mass of M?

37. anonymous

there is no answer on my textbook. I cannot ask my teacher since I will have Chemistry examination tonight.

38. sirm3d

it is the mass of M relative to the mass of O

39. anonymous

there are 1 atom of M, right?

40. 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

41. anonymous

so M is copper....

42. anonymous

mass of M=79.87g and the relative atomic mass is ...?

43. 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.

44. anonymous

how can you get the answer?

45. 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.

46. sirm3d

you cannot give the actual mass of M in the compound MO if you do not have the actual mass of MO.

47. 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?

48. 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...

49. 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.

50. sirm3d

16.0g O x 79.87g M / 20.13g O = 63.5g M

51. 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?

52. sirm3d

yep. that's how you convert the given.

53. 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

54. sirm3d

|dw:1354708533879:dw| observe how i got rid of g O and change it to g M

55. anonymous

I know it is the answer of part d 168 g M?

56. anonymous

are you here?

57. sirm3d

1 got 132.32 g M

58. anonymous

ummm......what are your steps?

59. sirm3d

g M in MCl = 268.88g-5.68g=21.2g M

60. sirm3d

mass ratio is 21.2g M / 5.68g Cl

61. anonymous

ummmm...you get something wrong.... is 26.88g instead of 268.88g

62. sirm3d

sorry. typo error.

63. sirm3d

relative mass of Cl = 35.453g

64. anonymous

just take the relative mass of Cl =35.5g

65. sirm3d

35.5g Cl x 21.2g M / 5.68g Cl = 132.5g M

66. sirm3d

M = Cs

67. anonymous

why 21.2 g?

68. anonymous

26.88g-5.68g?

69. anonymous

okay I got it. Thank you very much. :)

70. sirm3d

yep. 26.88g - 5.68g = 21.2g

71. 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.

72. sirm3d

hmm. 0.213 g O2 used in the process.

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