Answer:
C₅H₁₀O₄
Explanation:
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of C:H:O.
Assume 100 g of deoxyribose.
1. Calculate the mass of each element.
Then we have 44.8 g C, 7.5 g H, and 47.7 g O.
2. Calculate the moles of each element
![\text{Moles of C} = \text{44.8 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = \text{3.730 mol C}\\\\\text{Moles of H} = \text{7.5 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H }} = \text{7.44 mol H}\\\\\text{Moles of O} = \text{47.7 g O} \times \dfrac{\text{1 mol O}}{\text{16.00 g O }} = \text{2.981 mol O}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20C%7D%20%3D%20%5Ctext%7B44.8%20g%20C%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20C%7D%7D%7B%5Ctext%7B12.01%20g%20C%7D%7D%20%3D%20%5Ctext%7B3.730%20mol%20C%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20H%7D%20%3D%20%5Ctext%7B7.5%20g%20H%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20H%7D%7D%7B%5Ctext%7B1.008%20g%20H%20%7D%7D%20%3D%20%5Ctext%7B7.44%20mol%20H%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20O%7D%20%3D%20%5Ctext%7B47.7%20g%20O%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20O%7D%7D%7B%5Ctext%7B16.00%20g%20O%20%7D%7D%20%3D%20%5Ctext%7B2.981%20mol%20O%7D)
3. Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
C:H:O = 3.730:7.44:2.981 = 1.251:2.50:1 = 5.005:9.98:4 ≈ 5:10:4
4. Write the empirical formula
EF = C₅H₁₀O₄
How many moles of ammonia are produced when 6 moles of hydrogen gas react with nitrogen gas? What are some simple steps I can take to protect my privacy online? 6 * 2/3 moles of ammonia must be produced. Therefore 4 moles of ammonia can be produced
Answer:
Explanation:
All compounds are molecules, but not all molecules are compounds. That is because a molecule can be made up of two atoms of the same kind, as when two oxygen atoms bind together to make an oxygen molecule. However, all compounds are made up of two or more different types of atoms.
To determine the fraction of carbon in morphine, we need to know the chemical formula of morphine. From my readings, the chemical formula would be <span>C17H19NO<span>3. We assume we have 1 g of this substance. Using the molar mass, we can calculate for the moles of morphine. Then, from the formula we relate the amount of carbon in every mole of morphine. Lastly, we multiply the molar mass of carbon to obtain the mass of carbon. We calculate as follows:
1 g </span></span> <span>C17H19NO<span>3 ( 1 mol / 285.34 g ) ( 17 mol C / 1 mol </span></span> <span>C17H19NO3</span>) ( 12.01 g C / 1 mol C) = 0.7155 g C
Fraction of carbon = 0.7155 g C / 1 g <span>C17H19NO<span>3 = 0.7155</span></span>
D. 100%, extremely sure. Because Newton's 3rd law states every force has an equal and opposite reaction.