How would you do this? A 4.428 g sample of an oxide of Ru contains 3.578 g Ru. What is the empirical formula of the oxide?

Question

anonymous · Answer

Please someone help! :(

cuanchi · Answer

1) If your sample is 4.428g and 3.578g is Ru, the rest is oxygen 4.428 - 3.578= g of Oxygen 2) then divide the gRu (3.578g is Ru) by the atomic mass of Ru (look in a periodic table) and the g of O by the atomic mass of oxygen. 3) divide the both numbers obtained in 2) by the smallest of them, one quotient is going to be equal to one the other can be a whole number or a decimal number. If is a whole number just you can write the empirical formula as Ru"x" O"y". If is not a whole number you have to multiply both numbers by a same factor to convert both in whole numbers. https://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm