I need some help with some of these questions:

The action of bacteria on meat and fish produces a poisonous compound called cadaverine. As its name implies, it stinks! It is 58.77 percent carbon, 13.81 percent hydrogen, and 27.40 percent nitrogen. Its molecular weight is 102.2 g/mol. Determine the molecular formula.

Calculate the empirical formula for a chemical compound that gives the following analysis: 65.2 g scandium and 34.8 g oxygen.

A certain compound has an empirical formula of NH2O. Its molar mass is between 55 and 65 g/mol. What is its molecular formula?

compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45 percent C, 5.45 percent H, and 29.09 percent O. Determine the empirical formula and the molecular formula.

All of these are alike.

Convert g to moles. mols = g/molar mass.
Then find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself (giving it 1.00), then divide the other numbers by the same small number.
Post your work if you get stuck.

To determine the molecular formula, you can follow these steps:

1. Find the empirical formula by determining the ratios of the elements present in the compound.

For the compound with the percentage composition of 58.77% carbon, 13.81% hydrogen, and 27.40% nitrogen:
- Assume you have 100 grams of the compound.
- Convert the percentages to grams by multiplying each percentage by 100 grams.
- Convert the grams to moles by dividing each gram value by the molar mass of the respective element.
- Divide each mole value by the smallest mole value to get the mole ratios. If necessary, round the ratios to the nearest whole number to obtain whole numbers.
- The empirical formula is the ratio of the elements. In this case, it would be C3H8N.

2. Determine the molar mass of the empirical formula.

Calculate the molar mass of C3H8N by summing the molar masses of each element:
- C (carbon) has a molar mass of 12.01 g/mol.
- H (hydrogen) has a molar mass of 1.01 g/mol.
- N (nitrogen) has a molar mass of 14.01 g/mol.

Molar mass of C3H8N = (3 * 12.01) + (8 * 1.01) + 14.01 = 58.11 g/mol.

3. Divide the given molar mass by the molar mass of the empirical formula.

For the compound with a molar mass between 55 g/mol and 65 g/mol:
- Divide the given molar mass by the molar mass of the empirical formula: Molecular mass / Empirical mass = n (where n is an integer).
- In this case, the molecular mass is between 55 g/mol and 65 g/mol. Divide these values by 58.11 g/mol to find the values of n.
- The result will be a range of values for n.

4. Determine the molecular formula.

The molecular formula is obtained by multiplying the empirical formula by the value of n:
- For example, if n = 2, then the empirical formula NH2O becomes (NH2O)2 = N2H4O2.

Repeat the same steps for the compound with a molar mass of 100 g/mol and percentage composition of 65.45% C, 5.45% H, and 29.09% O.

To determine the molecular formula for each of these questions, you need to follow a series of steps. These steps involve calculating the empirical formula and then determining the molecular formula using the molar mass information.

1. Determining the molecular formula for cadaverine:
a. Start by converting the percentages of carbon, hydrogen, and nitrogen into grams. Assume you have 100g of the compound to simplify the calculations.
- Carbon: 58.77g (58.77% of 100g)
- Hydrogen: 13.81g (13.81% of 100g)
- Nitrogen: 27.40g (27.40% of 100g)

b. Convert the grams of each element into moles by dividing by their respective atomic masses:
- Carbon: 58.77g / 12.01 g/mol ≈ 4.89 mol
- Hydrogen: 13.81g / 1.01 g/mol ≈ 13.67 mol
- Nitrogen: 27.40g / 14.01 g/mol ≈ 1.96 mol

c. Divide each element's mole value by the smallest mole value to get the simplest whole-number ratio of atoms:
- Carbon: 4.89 mol / 1.96 mol ≈ 2.49
- Hydrogen: 13.67 mol / 1.96 mol ≈ 6.98
- Nitrogen: 1.96 mol / 1.96 mol = 1

d. Round the ratios to the nearest whole number (keep them as small as possible):
- Carbon: 2
- Hydrogen: 7
- Nitrogen: 1

e. This gives you the empirical formula: C2H7N.

2. Calculating the empirical formula for the compound with 65.2 g scandium and 34.8 g oxygen:
a. Convert the grams of each element into moles:
- Scandium: 65.2g / 44.96 g/mol ≈ 1.45 mol
- Oxygen: 34.8g / 16.00 g/mol ≈ 2.18 mol

b. Divide each element's mole value by the smallest mole value:
- Scandium: 1.45 mol / 1.45 mol = 1
- Oxygen: 2.18 mol / 1.45 mol ≈ 1.50

c. Round the ratios to the nearest whole number:
- Scandium: 1
- Oxygen: 2

d. This gives you the empirical formula: ScO2.

3. Determining the molecular formula for a compound with the empirical formula NH2O and molar mass between 55 g/mol and 65 g/mol:
a. Calculate the molar mass of the empirical formula:
- Nitrogen (N): 1 atom × 14.01 g/mol = 14.01 g/mol
- Hydrogen (H): 2 atoms × 1.01 g/mol = 2.02 g/mol
- Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol
Total molar mass of empirical formula NH2O: 14.01 g/mol + 2.02 g/mol + 16.00 g/mol = 32.03 g/mol

b. Divide the given molar mass range by the molar mass of the empirical formula to obtain a whole number multiple:
- Lower range: 55 g/mol ÷ 32.03 g/mol ≈ 1.72
- Upper range: 65 g/mol ÷ 32.03 g/mol ≈ 2.03

c. Multiply the empirical formula by the whole number multiple:
- Lower range: NH2O × 2 ≈ N2H4O2
- Upper range: NH2O × 2 ≈ N2H4O2

d. The molecular formula of the compound is N2H4O2.

4. Determining the empirical and molecular formulas for a compound with a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% O:
a. Convert the percentages to grams:
- Carbon: 65.45 g (65.45% of 100 g)
- Hydrogen: 5.45 g (5.45% of 100 g)
- Oxygen: 29.09 g (29.09% of 100 g)

b. Convert the grams of each element into moles:
- Carbon: 65.45 g / 12.01 g/mol ≈ 5.45 mol
- Hydrogen: 5.45 g / 1.01 g/mol ≈ 5.40 mol
- Oxygen: 29.09 g / 16.00 g/mol ≈ 1.82 mol

c. Divide each element's mole value by the smallest mole value:
- Carbon: 5.45 mol / 1.82 mol ≈ 3
- Hydrogen: 5.40 mol / 1.82 mol ≈ 3
- Oxygen: 1.82 mol / 1.82 mol = 1

d. Round the ratios to the nearest whole number:
- Carbon: 3
- Hydrogen: 3
- Oxygen: 1

e. This gives you the empirical formula: C3H3O.

f. Calculate the molar mass of the empirical formula:
- Carbon: 3 atoms × 12.01 g/mol = 36.03 g/mol
- Hydrogen: 3 atoms × 1.01 g/mol = 3.03 g/mol
- Oxygen: 1 atom × 16.00 g/mol = 16.00 g/mol
Total molar mass of the empirical formula C3H3O: 36.03 g/mol + 3.03 g/mol + 16.00 g/mol = 55.06 g/mol

g. Divide the molar mass of the given compound by the molar mass of the empirical formula:
- 100 g/mol ÷ 55.06 g/mol ≈ 1.82

h. Multiply the empirical formula by the whole number multiple:
- C3H3O × 1.82 ≈ C5H5O2

i. The empirical formula is C3H3O, and the molecular formula is C5H5O2.

Remember, these steps can be applied to solve similar problems in chemistry.