Can anyone help me figure out the formula for this hydrate?

Question

Can anyone help me figure out how to find the formula for this hydrate?

3.71 g Na2CO3, 6.29 g H2O

Thanks!

Answer 1

the empirical formula can be calculated;

1. Start with the number of grams of each element, given in the problem.
If percentages are given, assume that the total mass is 100 grams so that
the mass of each element = the percent given.

2. Convert the mass of each element to moles using the molar mass from the periodic table.

3. Divide each mole value by the smallest number of moles calculated.

4. Round to the nearest whole number. This is the mole ratio of the elements and is
represented by subscripts in the empirical formula.

If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same
factor to get the lowest whole number multiple.
e.g. If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.
e.g. If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.

Thus;

1. 3.71 g Na2CO3
6.29 g H2O

2. Mole conversion'

Use the formula n = m / M
n = moles
m = mass (g)
M = molar mass(g mol^-1)

n(Na2CO3) = 3.71 g / (23 + 23 + 12 + (16 x 3))
gmol^-1
n(Na2CO3) = 0.035mol

n(H2O) = 6.29g / (1 + 1 + 16) gmol^-1
n(H2O) = 0.349444444 mol

3. Divide each mole value by the smallest number of moles calculated.

In this case => n(Na2CO3) = 0.035mol

Na2CO3 = 0.035mol / 0.035mol =1
H2O = 0.349444444 mol / 0.035mol
H2O = 9.984126984


4. Na2CO3 = 1
H2O = 10

thus the empirical formula is Na2CO3.10H2O

Hope this helps:-)

Answer 2

Divide 3.71 by 106, and 6.29 by 18, and then divide the second answer by the first.