Answer:
2. All the naturally occurring isotopes of Mg.
Explanation:
You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.
1. and 3. are <em>wrong</em>. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.
4. is <em>wrong</em>. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)
Molecular mass of H2O is: 16+2*1=18
so moles of 40g of H2O is: 40/18=2.22
Answer:
'An ion has a non-zero electric charge. A radical has an atom with unfilled electron shells and so is very reactive, but is electrically neutral.'
'Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.'
'The primary difference that lies between these organic compounds and inorganic compounds is that organic compounds always have a carbon atom while most of the inorganic compounds do not contain the carbon atom in them.'
A contains 38.5 g of tin for each 12.3 g of fluorine:
<span>mole ratio: </span>
<span>(38.5 g)/(118.71 g/mol):(12.3 g)/(18.998 g/mol) = 0.324:0.647 = 1:2 ⇒ SnF₂ </span>
<span>B contains 56.5 g of tin for each 36.2 g of fluorine: </span>
<span>mole ratio: </span>
<span>(56.5 g)/(118.71 g/mol):(36.2 g)/(18.998 g/mol) = 0.476:1.905 = 1:4 ⇒ SnF₄
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