Answer
$$C_6H_9OBr$$
Work Step by Step
** Notice, if the compound had 2 bromines and 1 one of each other atom, it's molar mass would be:
M = 80.0 * 2 + 12.0 + 1.0 + 16.0 = 189 g/mol, which is greater than 177 g/mol.
Therefore, this molecule can't have 2 or more bromines, which leaves us to $n(Br) = 1$
$$C_xH_yO_zBr$$
$m(C)= 8m(H)$
$12.0 \times n(C) = 8 \times 1.0 \times n(H)$
$$12.0 x = 8 \times( 1.0 y)$$ $$\frac{12.0}{8.0} \ x= y$$ $$1.5 x = y$$
** The molar mass without the bromine is:
$177 \space g/mol - 80 \space g/mol = 97 \space g/mol$
$12.0 \space x + 1.0 \space y + 16.0 \space z = 97$
$12.0 \space x + 1.0 (1.5 \space x) + 16.0 \space z = 97$
$13.5 \space x + 16.0 \space z = 97$
$16.0 \space z = 97 - 13.5 \space x$
$$z = \frac{97 - 13.5 \space x}{16.0}$$
Let's try some values for 'x':
x = 2:
$$ z = \frac{97 - 13.5 (2)}{16.0} = 4.375$$
'z' is not an integer, so x=2 isn't valid.
x = 4:
$$ z = \frac{97 - 13.5 (4)}{16.0} = 2.6875$$
'z' is not an integer, so x=4 isn't valid.
x = 6
$$ z = \frac{97 - 13.5 (6)}{16.0} = 1$$
'z' is an integer, so x=6 is valid.
$$C_6H_{1.5(6)}OBr$$
$$C_6H_9OBr$$
