Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

Some Help with These Chem Problems Please? (Hydrated Crystals)?

1. A 6.88 g sample of the hydrated crystal Rb2CO3 • xH2O is heated until all the water is removed. The remaining solid weighs 5.24 g. Calculate the formula of the hydrated crystal

2.  A 4 g sample of the hydrated crystal Ba(OH)2 • xH2O is heated until all the water is removed. The remaining solid weighs 2.17 g. Calculate the formula of the hydrated crystal.

3. A 6.42 g sample of the hydrated crystal Sr(OH)2 • xH2O is heated until all the water is removed. The remaining solid weighs 4.04 g. Calculate the formula of the hydrated crystal.

4. What is the correct chemical formula for the hydrated crystal Mn(NO3)4 • xH2O if analysis shows 511 g of Mn(NO3)4 and 334 g of H2O?

5. What is the correct chemical formula for the hydrated crystal Li2SO4 • xH2O if analysis shows 75.6 g of Li2SO4 and 74.3 g of H2O?

Thanks

2 Answers

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  • 1 month ago
    Favorite Answer

    2.

    (4 g - 2.17 g) / (18.01532 g H2O/mol) = 0.10158 mol H2O

    (2.17 g Ba(OH)2) / (171.3417 g Ba(OH)2/mol) = 0.01266475 mol Ba(OH)2

    (0.10158 mol H2O) / (0.01266475 mol Ba(OH)2) = 8

    So the requested formula is:

    Ba(OH)2·8H2O

    3.

    (6.42 g - 4.04 g) / (18.01532 g H2O/mol) = 0.13210978 mol H2O

    (4.04 g Sr(OH)2) / (121.6347 g Sr(OH)2/mol) = 0.0332142 mol Sr(OH)2

    (0.13210978 mol H2O) / (0.0332142 mol Sr(OH)2) = 3.9775 = 4

    So the requested formula is:

    Sr(OH)2·4H2O

    4.

    (511 g Mn(NO3)4) / (302.9576 g Mn(NO3)4/mol) = 1.6867 mol Mn(NO3)4

    (334 g H2O) / (18.01532 g H2O/mol) = 18.5398 mol H2O

    (18.5398 mol H2O) / (1.6867 mol Mn(NO3)4) = 10.99 = 11

    So the requested formula is:

    Mn(NO3)4·11H2O

    5.

    (75.6 g Li2SO4) / (109.9446 g Li2SO4/mol) = 0.687619 mol Li2SO4

    (74.3 g H2O) / (18.01532 g H2O/mol) = 4.124268 mol H2O

    (4.124268 mol H2O) / (0.687619 mol Li2SO4) = 5.998 = 6

    So the requested formula is:

    Li2SO4·6H2O

  • 1 month ago

    I'll answer the first one.

    The general idea is to see how many moles of water are present when one mole of the anhydrous compound is present.

    6.88 - 5.24 = 1.64 g of water

    5.24 g / 230.944 g/mol = 0.0226895 mol

    1.64 g / 18.015 g/mol = 0.091035 mol

    230.944 g/mol is the molec. weight of Rb2CO3

    0.0226895 mol / 0.0226895 mol = 10.091035 mol / 0.0226895 mol = 4Rb2CO3 • 4H2O

    More examples:

    https://www.chemteam.info/Mole/Determine-formula-o...

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