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

F=121.61 gr,

Step-by-step explanation:

we see that it is a problem of proportion, so then if the total mass of the sample is the sum of M plus fluorine

the sample weighs 1.38gr M + 0.71gr F = 2.09, where M y F represent:

M=1.38*100/2.09 = 66.02% and

F=0.71*100/2.09 = 33.97%, also

in 358 gr. * 0.3397F = 121.61 gr F