Marine chemist exam in the mass percent composition of manganese sulfide compound from ocean floor he determined that 60.2% of this compound is sulfur determine the mole ratio between manganese and sulfur

Explanation:

Let the mass of the compound be 100g.

There are 60.2g of sulfur.

Moles of sulfur

= 60.2g / (32.07g/mol) = 1.877mol

There are 100g - 60.2g

= 39.8g of manganese.

Moles of manganese

= 39.8g / (54.94g/mol) = 0.724mol

Mole ratio of manganese to sulfur

= 0.724mol : 1.877mol = 1 : 2.5.

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.