The mass percentage of bismuth in the compound will be equal to the mass of bismuth in one mole of compound divided by the total molar mass of the compound.

Bismuth has a molar mass of $\displaystyle 208.98\frac{g}{mol}$. One mole of the compound would result in two moles of bismuth, a total of 417.96g.

$\displaystyle 2(208.98)=417.96g$

Tellurium has a molar mass of $\displaystyle 127.6\frac{g}{mol}$. One mole of the compound would result in three moles of tellurium, a total of 382.8g.

$\displaystyle 3(127.6)=382.8g$

Add the mass of bismuth and the mass of tellurium per mole to find the total molar mass.

$\displaystyle MM=417.96g+382.8g=800.76g$

Divide the mass of bismuth by the total molecular mass to find the percent by mass of bismuth in the compound.

$\displaystyle \frac{417.96g}{800.76g}\times100\%=52.2\%$

The total molar mass of acetic acid is $\displaystyle 60\frac{g}{mol}$.

$\displaystyle 2(12)+2(16)+4(1)=60\frac{g}{mol}$

Carbon contributes $\displaystyle 24g$ total. Therefore, in a sample of 1 mole of acetic acid, there are $\displaystyle 24g$ of carbon. To find the percent by mass, we must divide the mass of carbon by the total molar mass of the compound.

$\displaystyle \frac{24g}{60g}=40\%$

The total molar mass of lead (II) sulfate is $\displaystyle 303.265g/mol$. Lead contributes $\displaystyle 207.2g$, sulfur contributes $\displaystyle 32.065g$, and oxygen contributes $\displaystyle (4*16g)=64g$.

The percent by mass of each element in the compound is found by dividing the mass contribution of that element by the total molar mass of the compound.

$\displaystyle Pb: \frac{207.2gPb}{303.265gtotal}=68.33\%$

$\displaystyle S: \frac{32.065gS}{303.265gtotal}=10.57\%$

$\displaystyle O: \frac{64gO}{303.265gtotal}=21.1\%$

The molar mass of calcium chloride is $\displaystyle 110.987\frac{g}{mol}$. Chlorine has two moles per one mole of $\displaystyle CaCl_{2}}$. Therefore, chloride contributes 70.9g to the total molecular mass.

$\displaystyle (2mol*35.45\frac{g}{mol})=70.9g$

To find the percent by mass of chlorine in calcium chloride, divide the contribution of chlorine by the total molar mass.

$\displaystyle \frac{70.9g}{110.987g}=63.9\%$

The total mass of one mole of aluminum (II) chromate is calculated by:

$\displaystyle 2(26.98\frac{g}{mol} Al)+3(51.99\frac{g}{mol} Cr)+12(16\frac{g}{mol}O)=401.95\frac{g}{mol}Al_2(CrO_4)_3$

Aluminum, chromate, and oxygen contribute $\displaystyle 53.96g, 155.97g,$ and $\displaystyle 192g$respectively. Therefore, we can divide each contribution by the total molecular mass to determine percentages by mass.

$\displaystyle Al: \frac{53.96gAl}{401.95gtotal}=13.42\%$

$\displaystyle Cr: \frac{155.97gCr}{401.95gtotal}=38.8\%$

$\displaystyle O: \frac{192gO}{401.95gtotal}=47.78\%$

The total mass of $\displaystyle U(C_{2}H_{3}O_{2})_{3}$ is $\displaystyle 415.16\frac{g}{mol}$ and calculated by:

$\displaystyle (238.0289\frac{g}{mol}U)+6(12.01\frac{g}{mol}C)+9(1.008\frac{g}{mol}H)+6(16\frac{g}{mol}O)=415.16\frac{g}{mol}U(C_2H_3O_2)_3$

You can then divide hydrogen's contribution by the total molecular mass in order to find the percent mass.

$\displaystyle \frac{9.072g}{415.16g}=2.18\%$

The total molar mass of $\displaystyle Cs_{2}Se$ is $\displaystyle 344.78\frac{g}{mol}$. Cesium has a +1 charge when the compound is dissolved in solution and is, thus, the cation.

$\displaystyle Ce_2Se\rightarrow2Ce^{+1}+Se^{-2}$

Cesium contributes 265.8g to the total molecular mass.

$\displaystyle (2mol*132.9\frac{g}{mol}Cs)=265.8gCs$

We can find the percent by mass of cesium by dividing the mass of cesium by the total molecular mass.

$\displaystyle \frac{265.8gCs}{344.78g\ total}=77.1\%$

When assigning oxidation states to elements of a given compound, non-transition metal elements are assigned specific oxidation states corresponding to their group number and valence relative to a complete octet.

Group 1 elements have 1 valence electron and an oxidation state of +1.

Group 2 elements have 2 valence electrons and an oxidation state of +2. 

Group 8 elements (the noble gases) have a complete octet, thus their assigned oxidation state is 0.

Group 7 elements (halogens) have 7 valence electrons and an oxidation state of -1.

Group 6 elements such as oxygen have 6 valence electrons and an oxidation state of -2.

Permanganate has 4 oxygen atoms and an overall charge of -1. The oxidation state of the $\displaystyle Mn$ atom may be found by first calculating the combined oxidation state of the oxygen atoms:

$\displaystyle 4 * (-2) = -8$ 

and finding the difference between their combined oxidation state and the overall charge of the ion (-1):

$\displaystyle (-1) - (-8) = +7$

Both the electron-pair and molecular geometries of ammonium are tetrahedral, as opposed to ammonia ($\displaystyle NH_3$), in which the electron-pair geometry is tetrahedral, but the molecular geometry is trigonal pyramidal. As evidenced below:

Ammonia:

Note that the difference in geometry arises in the presence of the lone pair on nitrogen in ammonia. The lone pair only contributes to molecular geometry, but does not contribute to electron-pair geometry. Bonding electrons, however, contribute to both electron and molecular geometries.

Ammonium:

In ammonium, the lone pair seen in ammonia is shared in a bond with an additional hydrogen atom. 

On average, the molar masses of carbon, hydrogen, and oxygen are approximately $\displaystyle 12\frac{g}{mol}$,  $\displaystyle 1\frac{g}{mol}$, and $\displaystyle 16\frac{g}{mol}$, respectively.

The molar mass of a molecule of glucose is:

$\displaystyle \left (6 * 12 \frac{g}{mol} \right ) + \left (12 * 1 \frac{g}{mol} \right ) + \left (6 * 16 \frac{g}{mol} \right ) = 180 \frac{g}{mol}$ 

Each term added in the previous step is the mass of each element present in a mole of glucose, so the mass percent of its components are found by dividing their respective contributions to the molar mass by the molar mass:

$\displaystyle C: \frac{72\frac{g}{mol}}{180\frac{g}{mol}} * 100 \% = 40\%$

$\displaystyle H: \frac{12\frac{g}{mol}}{180\frac{g}{mol}} * 100 \% = 6.7\%$

$\displaystyle O: \frac{96\frac{g}{mol}}{180\frac{g}{mol}} * 100\% = 53.3\%$