Elemental Composition of Pure Substances - AP Chemistry

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.

The molecular weight of K2Cr2O7 is 2(39 g/mol) + 2(52 g/mol) + 7(16 g/mol) = 294 g/mol.

Out of this total weight, the percent composition of K is: (2(39 g/mol)/294 g) = 26.5%

An empirical formula represents a molecule with the simplest ratio. C6H6, ethyene (C2H2) and C100H100 can all be reduced to CH, however, C3H4 cannot because it does not have C and H atoms in a 1 : 1 ratio.

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%.

The empirical formula of a compound is the simplest whole number ratio of the elements in the compound, whereas the molecular formula gives the exact number of atoms of each element in the compound—this is usually a multiple of the empirical formula. If the empirical formula of a compound is CH, a common molecular formula is C6H6 (benzene).

Cobalt (II) is Cobalt in its second oxidation state, , and the chemical formula for Nitrate is . The neutral salt is .

Ferrous is the name for Iron (II), or . Sulfide is the anion of sulfur, . The chemical formula for ferrous sulfide is .

is a saturated hydrocarbon with an organic functional group, , or alcohol.

is propane, so we will remove the "e" from the end of that name and add the suffix "ol" to denotate an alcohol, so the correct answer is propanol.

Propenol is propanol with a double bond in the hydrocarbon chain. Ethanol is a saturated two-carbon chain with an alcohol functional group.

The correct answer is . We found that element Y is aluminum due to its discrepancy between IE3 and IE4. Aluminum contains three valence electrons, and it needs to shed all three to reach a noble gas configuration. This makes it Al(III), or Al3+. Oxygen must gain two electrons to achieve a noble gas configuration. This makes it O2-. The only way to balance the opposing charges of the aluminum oxide is to have two aluminum ionically bonded to three oxygen.

Aluminum: charge +3; Oxygen charge: -2

2(+3) + 3(-2) = 0

If we assume that we have a sample of 100g of the compound, then that would mean we have 37.47 grams (3.12 moles) of carbon, 12.61 grams (12.48 moles) of hydrogen, and 49.92 grams (3.12 moles) of oxygen.

The molar ratio of C:H:O is 3:12:3, reducing to 1:4:1, giving the empirical formula of CH4O.

Remember that sodium wants to lose only one electron in order to achieve a stable octet. Meanwhile, oxygen needs to gain 2 electrons in order to achieve a stable octet. As a result, the expected compound involves two sodium atoms, each giving up one electron in order to fill oxygen's octet.

Covalent non-metal/non-metal compounds are named by taking the name of each element and appending the apropriate greek prefix to indicate the number of each atom present—in this case, di- for two and tetra- for four. The end of the more electronegative element is substituted with "ide". This gives the name of dinitrogen tetraoxide.

In order to find the molecular formula, we must first find the empirical formula. We start by imagining a sample of the compound weighing 100 grams, so the percentages can be seen as grams. 25g of the sample is carbon, 8.3g of the sample is hydrogen, and 66.7g of the sample is oxygen.

Then, each of the atoms' masses is divided by the atom's molar mass, the numbers are then compared to each other in order to find the ratio of atoms.

There is a ratio of two carbon to eight hydrogen to four oxygen, which reduces to a 1:4:2 ratio between carbon, hydrogen, and oxygen respectively.

As a result, the empirical formula is . Since the molecular formula is three times as heavy, we simply multiply each portion of the empirical formula by three. The molecular formula is .

To solve this problem, we will consider a sample that is 100g. We can assume that 75g of that sample is carbon and 25g of that sample is hydrogen.

First, we will determine the molar ratio of carbon to hydrogen.

Now we will determine the whole number ratio.

In , the nitrogen is bound covalently to the three oxygens, and the complex has an overall of .

Sodium has a charge of , and is ionically bound to the complex.

When put into water, the compound will dissociate into and .

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

We should first remember the difference between sulfate, sulfite, and sulfide. Sulfate is , sulfite is ` and sulfide is .

The only answer choices that could be right must have in them. We then need to see that barium usually has a charge of , as the periodic table shows us, and so we need a charge of to cancel that out. The answer is .

Alkaline earth metals are in group II of the periodic table. Of the given answer options, strontium (Sr) is the only element in group II.

Rubidium (Rb) is an alkali metal (group I). Iron (Fe) is a transition metal. Iodine (I) and bromine (Br) are halogens.

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.

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.

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.

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.

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

This results in a molecular formula of .

When finding empirical formulas, take the percentages of each atom and convert them to grams by imagining a 100-gram sample of the compound.

Next, divide each mass by the molar mass of the atom.

Finally, divide each number by the smallest number out of the three. Remember that ratios must be a whole number.

Since dividing the number for oxygen by the smallest number gives a value of 2.5, we must multiply each atom by 2 in the empirical formula.

This results in an empirical formula of .