Empirical formula is the simplest ratio of whole numbers of components in a compound
in 100 g of compound
C H O
mass 25.5 g 6.40 g 68.1 g
number of moles 25.5 g/12 g/mol 6.40 g/ 1 g/mol 68.1 g/ 16 g/mol
= 2.13 mol = 6.40 mol = 4.26 mol
divide by least number of moles
2.13/2.13 = 1 6.40/2.13 = 3.0 4.26/2.13 = 2.0
all rounded off
C - 1
H - 3
O - 2

empirical formula - CH₃O₂

8 0

3 years ago

How many moles are in 7.00E24 atoms of Ca?

iris [78.8K]

Explanation:

In chemistry, a mole is the base unit of the amount of substance in a system containing elementary entities (atoms, molecules, ions, electrons), and one mole of substance corresponds to the Avogadro's constant or $6.023 \ \times \ 10^{23}$ entities. The stoichiometry equation that relates the amount of substance into the measurement in moles is

$n \ = \ \displaystyle\frac{\alpha}{N_{A}}$ ,

where $n$ is the number of moles, $\alpha$ is the amount of substance and $N_{A}$ is the Avogadro's constant.

Therefore,

$n \ = \ \displaystyle\frac{7.00 \ \times \ 10^{24} \ \text{atoms}}{6.023 \ \times \ 10^{23} \ \text{atoms mol}^{-1}} \\ \\ n \ = \ 11.624 \ \text{mol}$ .

3 0

2 years ago

If 10.g of AgNo3 is available, what volume of 0.25 M AgNo3 can be prepared

Marianna [84]

The equation is L = m/M
First, covert 10. grams of AgNO3 to moles which is 0.059 moles.
Divide 0.059 moles by 0.25M which is 0.24 liters.

3 0

3 years ago