name the compound which contains 0.463g Ti (#81), 0.0544g of carbon,0.00685g of hydrogen and0 0.0725g of oxygen by finding its empirical formula
To determine the empirical formula, we need to find the moles of each element present in the compound.
First, let's find the moles of each element present:
- Moles of Ti: (0.463 g Ti) / (47.867 g/mol) = 0.00966 mol Ti
- Moles of C: (0.0544 g C) / (12.011 g/mol) = 0.00453 mol C
- Moles of H: (0.00685 g H) / (1.00784 g/mol) = 0.00680 mol H
- Moles of O: (0.0725 g O) / (15.999 g/mol) = 0.00453 mol O
Next, we need to divide each mole value by the smallest mole value to obtain the simplest whole number ratio:
- Moles of Ti: 0.00966 mol Ti / 0.00453 mol C ≈ 2.135
- Moles of C: 0.00453 mol C / 0.00453 mol C = 1
- Moles of H: 0.00680 mol H / 0.00453 mol C ≈ 1.5
- Moles of O: 0.00453 mol O / 0.00453 mol C = 1
The empirical formula is therefore Ti2C1.5O1 (rounded to the nearest whole number).