We are given a mass ratio of the elemental components, so all we have to do is divide each given mass by the respective atomic mass of its element. Approximately, the atomic masses of the elements in question are:

$\displaystyle C=\frac{360}{12} = 30$

$\displaystyle H=\frac{40}{1} = 40$

$\displaystyle N=\frac{280}{14} = 20$

$\displaystyle O=\frac{320}{16} = 20$

We now have to reduce this proportion to its smallest possible terms. Since they are all visibly multiples of 10, it makes sense to divide them by 10, yielding:

$\displaystyle C=\frac{30}{10}=3$

$\displaystyle H=\frac{40}{10}=4$

$\displaystyle N=\frac{20}{10}=2$

$\displaystyle O=\frac{20}{10}=2$

Since there are 3 carbons, and 3 is a prime number, this is the smallest whole number proportion we can have. Plug the numbers into a formula as subscripts, yielding the empirical formula:

$\displaystyle C_3H_4N_2O_2$

The molecular formula for glucose is $\displaystyle C_6H_{12}O_6$. The empirical formula is the most reduced formula. So think of the biggest number that: 6, 12, and 6 can be divided by. All of those numbers can be divided by 6, so we end up with 1, 2, 1 as our coefficients. That means we should have 1 carbon, 2 hydrogens, and 1 oxygen in the most reduced formula. Note that the 1:2:1 ratio of these elements is characteristic of all carbohydrates. They are hydrates (water) of carbon. Thus they contain one water molecule $\displaystyle (H_2O)$ for each carbon molecule.

In this question, we're given the elements that make up a given compound, and we're told the relative percentages of each element. We're asked to determine the empirical formula.

Though there are a few different ways to approach this problem, perhaps the easiest one is to first imagine that we're starting off with $\displaystyle 100\:grams$ of the compound. This way, we can directly convert the percentage of each element to grams. After we do this, we can convert the amount of grams of each element into the number of moles for that element.

$\displaystyle 39\:grams\:S\: (\frac{1\: mol\: S}{32.06\: grams\: S})=1.216\:moles\:S$

$\displaystyle 58.6\:grams\:O\: (\frac{1\: mol\: O}{16\: grams\: O})=3.663\:moles\:O$

$\displaystyle 2.4\:grams\:H\: (\frac{1\: mol\: H}{1.008\: grams\: H})=2.381\:moles\:H$

Now that we have the number of moles of each element, we can list our compound as follows.

$\displaystyle S_{1.216}O_{3.663}H_{2.381}$

Note that this is not yet the empirical formula of our compound, because an empirical formula has the simplest integer number for each of its elements. To arrive at our empirical formula, we can divide each element's number by the lowest number among them. In this case, sulfur's number is the lowest.

$\displaystyle S_{\frac{1.216}{1.216}}O_{\frac{3.663}{1.216}}H_{\frac{2.381}{1.216}}=S_{1}O_{3.012}H_{1.958}$

The numbers above are very close to whole number integers, so we can safely round them off. Thus, the empirical formula that we arrive at for this compound is the following.

$\displaystyle SO_{3}H_{2}$

When finding molecular formulas, imagine a 100-gram sample of the compound. We can then use the percentages of each atom and convert them to grams.

$\displaystyle 100g\ total$

$\displaystyle 13g\ C,\ 4.3g\ H,\ 30.4g\ N,\ 52.2g\ O$

The next step is to divide the given mass of each atom by its atomic mass. This will give you four separate molar values that you can compare to one another.

$\displaystyle 13g\ C*\frac{1mol}{12g}=1.08mol\ C$

$\displaystyle 4.3g\ H*\frac{1mol}{1.0g}=4.3mol\ H$

$\displaystyle 30.4g\ N*\frac{1mol}{14g}=2.17mol\ N$

$\displaystyle 52.2g\ O*\frac{1mol}{16g}=3.26mol\ O$

Next, you must divide each molar value by the smallest of the values. This will result in the molar ratios of each compound. In this case, carbon has the lowest molar value at 1.08. After dividing all four numbers by this value, we determine a molar ratio of 1:4:2:3 for carbon, hydrogen, nitrogen, and oxygen, respectively.

$\displaystyle \frac{1.08}{1.08}:\frac{4.3}{1.08}:\frac{2.17}{1.08}:\frac{3.26}{1.08}$

$\displaystyle 1:4:2:3$

$\displaystyle CH_4N_2O_3$

At this point, we have determined the empirical formula for the compound, however, we need to find the formula that gives a molar mass of 184 grams per mole. Start by calculating the molar mass of the empirical formula.

$\displaystyle 12(C)+4(H)+2(N)+3(O)$

$\displaystyle 12(1)+4(1)+2(14)+3(16)=92\frac{g}{mol}$

Since the empirical formula has a molar mass of 92 grams per mole, we need to multiply the empirical formula by 2.

$\displaystyle 2*92\frac{g}{mol}=184\frac{g}{mol}$

This results in a molecular formula of $\displaystyle \small C_{2}H_{8}N_{4}O_{6}$.

Here we are looking for the molecular formula of an ionic compound consisting of Ca and OH. We know that the molecule is neutral, so it must have a charge of 0.  This question is really addressing one's knowledge of typical ionic species.  Calcium ions are found only as Ca²⁺ and hydroxide ions (OH) are found only as OH–.  Thus for these two species to exist in an neutral ionic compound the only possible formula is Ca(OH)_2.  

To find the correct formula, the charge contributions from each compound must cancel out to zero. Each magnesium ion has a charge of $\displaystyle \small +2$ and each nitrogen ion has a charge of $\displaystyle \small -3$, due to their respective valence electron configurations. Three magnesium atoms would produce a charge of $\displaystyle \small +6$, and two nitrogen atoms would produce a charge of $\displaystyle \small {-6}$. This would thus create a neutral, stable compound.

$\displaystyle 3(2)+2(-3)=0$

$\displaystyle 3(Mg)+2(N)=0$

$\displaystyle Mg_3N_2$

To find the correct formula, the charge contributions from each compound must cancel out to zero. Each lithium ion has a charge of $\displaystyle \small +1$ and each oxygen ion has a charge of $\displaystyle \small -2$, due to their respective valence electron configurations. Two lithium atoms would produce a charge of $\displaystyle \small +2$, and one oxygen atom would produce a charge of $\displaystyle \small {-2}$. This would thus create a neutral, stable compound.

$\displaystyle 2(1)+1(-2)=0$

$\displaystyle 2(Li)+1(O)=0$

$\displaystyle Li_2O$

Acetone, which as a formula of C₃H₆O has a total molecular weight of: 3(12) + 6(1) + 1(16) = 58g/mole, and the percent of this that carbon makes up is (3(12)/58) X100 = 63%. 

When finding the formula for a neutral compound, you must make sure that the positive charges on the cation are cancelled out by the negative charges on the anion. Iron is a transitional element, and can create a couple of different cations depending on how many electrons it loses. The roman numeral III tells you the charge on the iron cation is 3+ ($\displaystyle \small Fe^{3+}$). Oxygen as an anion carries a charge of -2 ($\displaystyle O^{2-}$).  To make the compound neutral, the charges from oxygen must equal the charges from iron.

$\displaystyle 3(Fe)+(-2)(O)=0$

$\displaystyle 3Fe=2O$

$\displaystyle \frac{3}{2}=\frac{O}{Fe}$

This solution tells us the ratio of oxygen to iron: for every three oxygen, there are two iron.

$\displaystyle \small Fe_{2}O_{3}$ gives a total cation charge of +6 (2 iron ions), and a total anion charge of -6 (3 oxygen anions). These charges offset each other, resulting in a neutral compound.

In $\displaystyle NaNO_3$, the nitrogen is bound covalently to the three oxygens, and the $\displaystyle NO_3$ complex has an overall of $\displaystyle -1$.

Sodium has a charge of $\displaystyle +1$, and is ionically bound to the $\displaystyle NO_3$ complex.

When put into water, the compound will dissociate into $\displaystyle Na^+$ and $\displaystyle (NO_3)^-$.

HCl is an ionic compound, while the other answer choices have only covalent bonds.