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kryton1212

  • 2 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. kryton1212
    • 2 years ago
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    @gerryliyana

  2. gerryliyana
    • 2 years ago
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    (Ar Nitrogen/Mr NaNo3) x mass of NaNO3

  3. kryton1212
    • 2 years ago
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    Ar stands for what?

  4. gerryliyana
    • 2 years ago
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    mass of relative atom

  5. gerryliyana
    • 2 years ago
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    Mr is "mass of relative molecule"

  6. kryton1212
    • 2 years ago
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    ok

  7. kryton1212
    • 2 years ago
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    no need to times 100g?

  8. gerryliyana
    • 2 years ago
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    do you have a periodic table?

  9. kryton1212
    • 2 years ago
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    yes

  10. kryton1212
    • 2 years ago
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    wait a moment

  11. sirm3d
    • 2 years ago
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    \[\large m \text { N}=100g \text { Na} \times \frac{ 14g \text{ N} }{ 23g \text { Na} } \approx 61g \text { N}\]

  12. kryton1212
    • 2 years ago
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    Na=23.0g N=14.0g O=16.0g

  13. kryton1212
    • 2 years ago
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    why 23g Na?? not NaNO3?

  14. kryton1212
    • 2 years ago
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    but my idea is this... |dw:1354705910345:dw|

  15. sirm3d
    • 2 years ago
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    our interest is in the element Na and N in the compound. the ratio of these elements is 1:1

  16. kryton1212
    • 2 years ago
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    uhhh...what do you mean?

  17. sirm3d
    • 2 years ago
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    the compound contains 100g of Na. it did not say the compound NaNO3 weighs 100g.

  18. gerryliyana
    • 2 years ago
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    yes sodium is Na.., so 100 is mass of Na

  19. kryton1212
    • 2 years ago
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    @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
    • 2 years ago
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    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. gerryliyana
    • 2 years ago
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    represent mass of Na

  22. kryton1212
    • 2 years ago
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    @sirm3d okay, how about part b? @gerryliyana thanks, I got it :) part b please

  23. kryton1212
    • 2 years ago
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    4.6*(10H2O/Na2) ?

  24. sirm3d
    • 2 years ago
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    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
    • 2 years ago
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    mass of 10 moles of water is 10 x 18g = 180 g

  26. sirm3d
    • 2 years ago
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    we convert 4.6 g Na to g H2O by the mass ratio in the compound (180 g H2O / 23 g Na)

  27. kryton1212
    • 2 years ago
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    but there are two Na atom.... Na2Co3

  28. kryton1212
    • 2 years ago
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    sorry for typing mistakes, it should be Na2CO3 10H2O

  29. sirm3d
    • 2 years ago
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    sorry, missed that two Na. Mass of Na in compound is 46 g. use the mass ratio 180g H2O / 46g Na

  30. kryton1212
    • 2 years ago
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    |dw:1354706692310:dw|

  31. sirm3d
    • 2 years ago
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    right.

  32. kryton1212
    • 2 years ago
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    thanks. How about part c?

  33. sirm3d
    • 2 years ago
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    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. kryton1212
    • 2 years ago
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    I got this one

  35. sirm3d
    • 2 years ago
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    i got 63.5g M

  36. kryton1212
    • 2 years ago
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    so what is the formula about finding relative atomic mass of M?

  37. kryton1212
    • 2 years ago
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    there is no answer on my textbook. I cannot ask my teacher since I will have Chemistry examination tonight.

  38. sirm3d
    • 2 years ago
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    it is the mass of M relative to the mass of O

  39. kryton1212
    • 2 years ago
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    there are 1 atom of M, right?

  40. sirm3d
    • 2 years ago
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    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. kryton1212
    • 2 years ago
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    so M is copper....

  42. kryton1212
    • 2 years ago
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    mass of M=79.87g and the relative atomic mass is ...?

  43. sirm3d
    • 2 years ago
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    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. kryton1212
    • 2 years ago
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    how can you get the answer?

  45. sirm3d
    • 2 years ago
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    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
    • 2 years ago
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    you cannot give the actual mass of M in the compound MO if you do not have the actual mass of MO.

  47. sirm3d
    • 2 years ago
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    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. kryton1212
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    16.0g O x 79.87g M / 20.13g O = 63.5g M

  51. kryton1212
    • 2 years ago
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    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
    • 2 years ago
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    yep. that's how you convert the given.

  53. kryton1212
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1354708533879:dw| observe how i got rid of g O and change it to g M

  55. kryton1212
    • 2 years ago
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    I know it is the answer of part d 168 g M?

  56. kryton1212
    • 2 years ago
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    are you here?

  57. sirm3d
    • 2 years ago
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    1 got 132.32 g M

  58. kryton1212
    • 2 years ago
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    ummm......what are your steps?

  59. sirm3d
    • 2 years ago
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    g M in MCl = 268.88g-5.68g=21.2g M

  60. sirm3d
    • 2 years ago
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    mass ratio is 21.2g M / 5.68g Cl

  61. kryton1212
    • 2 years ago
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    ummmm...you get something wrong.... is 26.88g instead of 268.88g

  62. sirm3d
    • 2 years ago
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    sorry. typo error.

  63. sirm3d
    • 2 years ago
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    relative mass of Cl = 35.453g

  64. kryton1212
    • 2 years ago
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    just take the relative mass of Cl =35.5g

  65. sirm3d
    • 2 years ago
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    35.5g Cl x 21.2g M / 5.68g Cl = 132.5g M

  66. sirm3d
    • 2 years ago
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    M = Cs

  67. kryton1212
    • 2 years ago
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    why 21.2 g?

  68. kryton1212
    • 2 years ago
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    26.88g-5.68g?

  69. kryton1212
    • 2 years ago
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    okay I got it. Thank you very much. :)

  70. sirm3d
    • 2 years ago
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    yep. 26.88g - 5.68g = 21.2g

  71. kryton1212
    • 2 years ago
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    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
    • 2 years ago
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    hmm. 0.213 g O2 used in the process.

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