A mineral contains only a metal M and fluorine. If analysis indicates that a sample of the mineral contains 1.38 g of M and 0.71 g of F, what mass of fluorine is present in 358 g of the mineral? Enter your answer in decimal format with two decimal places and no units.

121.61

Step-by-step explanation:

The law of definite proportion sates that all pure samples of a particular chemical compound contain similar elements combined in the same proportion by mass.

From the question above,

The pure sample contains only metal M and Fluorine F.

Ratio of the mass of  M and F in the first sample = 1.38 g : 0.71 g

Total mass = 1.38+0.71 = 2.09 g

Percentage of F present in the first sample = (0.71/2.09)×100

= 33.97%

Assuming both samples are pure,

Total mass in the second sample = 358 g.

mass of fluorine = (33.97/100)×358

mass of fluorine = 121.61